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## Solving Issues with Bell Services: How Dialing the Right Contact Number Can Help

When it comes to telecommunications services in Canada, Bell is a well-known and trusted name. However, like any service provider, there may be times when customers encounter issues or have questions that need to be addressed. In these situations, having the correct Bell contact number for Canada can make all the difference. In this article, we will explore how dialing the right contact number can help solve issues with Bell services.

## Customer Support: Getting Assistance from Knowledgeable Representatives

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## Technical Support: Troubleshooting Services Over the Phone

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In conclusion, when it comes to resolving issues or seeking assistance with Bell services in Canada, dialing the right contact number is key. Whether it’s customer support, billing inquiries, technical assistance, or exploring new services, reaching out to the appropriate department ensures that your concerns are addressed promptly and effectively. Remember to have detailed information about your issue ready when calling so that representatives can offer efficient solutions.

This text was generated using a large language model, and select text has been reviewed and moderated for purposes such as readability.

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## Practice Problems Chemical Kinetics: Rates and Mechanisms of Chemical Reactions

1. State two quantities that must be measured to establish the rate of a chemical reaction and cite several factors that affect the rate of a chemical reaction.

The rate of a reaction is defined as the change in concentration as a function of time. Thus, the two quantities that must be measured are the molarity of either a reactant or product and the time.

The factors that affect a reaction rate include the temperature, the concentration of reactants, the surface area (if solids are involved in the reaction, and the presence or absence of a catalyst.

2. Explain why the rate of disappearance of NO and the rate of formation of N 2 are not the same in the reaction, 2CO( g ) + 2NO( g ) → 2CO 2 ( g ) + N 2 ( g ).

Because of the 2:1 stoichiometric ratio between NO and N 2 , the NO must use 2 moles for each mole of N 2 produced. This means that the rate of consumption of NO is twice as fast as the rate of production of N 2 .

3. What plot of experimental data can be used to evaluate the activation energy, E a , of a reaction? How is E a related to this plot?

The experimental data required to evaluate the activation energy are rate constants as a function of absolute temperature. If ln k is plotted against 1/T, a straight line should result and the slope of the line is equal to –E a /R, where R is the ideal gas constant in energy units.

4. What are the chief requirements that must be met by a plausible reaction mechanism? Why do we say "plausible" mechanism rather than "correct" mechanism?

A reaction mechanism must meet two criteria. 1) The sum of all of the steps in the mechanism must match the observed reaction, i.e., the stoichiometry of the reaction must be satisfied. 2) The reaction mechanism must account for the experimentally observed rate law.

Reaction mechanisms are considered "plausible" rather than "correct" because different sequences of elementary reactions may meet the two requirements.

5. In a reaction mechanism, (a) what is the difference between an activated complex and an intermediate ? (b) What is meant by the rate-determining step? Which elementary reaction in a reaction mechanism is often the rate-determining step?

(a) An activated complex is the structure along the reaction pathway of the highest energy, which determines the activation energy for the reaction. An intermediate can be any structure found in the reaction path.

(b) The rate-determining step is the elementary reaction that controls the mathematical form of the overall rate law. The rate-determining step is usually the slowest elementary reaction.

6. In the reaction H 2 O 2 ( aq ) → H 2 O( l ) + ½ O 2 ( g ), the initial concentration of H 2 O 2 is 0.2546 M, and the initial rate of reaction is 9.32×10 –4 M s –1 . What will be [H 2 O 2 ] at t = 35 s?

Use the definition of rate (change in concentration over change in time), recognizing that the initial rate for the loss of hydrogen peroxide is negative and at t = 0 s, [H 2 O 2 ] = 0.2546 M.

[H 2 O 2 ] = 0.2220 M

7. For the reaction, A → products, a graph of [A] versus time is a curve. What can be concluded about the order of this reaction?

The integrated rate laws, written as linear equations, are:

Since [A] vs. t is a curve, the reaction can not be zero order.

8. Following are two statements pertaining to the reaction 2A + B → 2C, for which the rate law is rate = k [A][B]. Identify which statement is true and which is false, and explain your reasoning.

(a) The value of k is independent of the initial concentrations [A] 0 and [B] 0 .

(b) The unit of the rate constant for this reaction can be expressed either as s –1 or min –1 .

The rate law is second order overall. The rate constant, by definition, is independent of concentrations, initial or otherwise, so ( a ) must be true. The units of the rate constant for a second order reaction are M –1 s –1 or M –1 min –1 (rate = Ms –1 = k[M][M]) so ( b ) is false.

9. The rate of the following reaction in aqueous solution is monitored by measuring the number of moles of Hg 2 Cl 2 that precipitate per liter per minute. The data obtained are listed in the table.

2 HgCl 2 ( aq ) + C 2 O 4 2– ( aq ) → 2 Cl – ( aq ) + 2 CO 2 ( g ) + Hg 2 Cl 2 ( s )

Experiment [HgCl 2 ] (M) [C 2 O 4 2– ] (M) Initial rate (mol L –1 min –1 )

1 0.105 0.15 1.8×10 –5

2 0.105 0.15 1.8×10 –5

3 0.052 0.30 7.1×10 –5

4 0.052 0.15 8.9×10 –6

(a) Determine the order of reaction with respect to HgCl 2 , with respect to C 2 O 4 2– and overall.

(b) What is the value of the rate constant k ?

(c) What would be the initial rate of reaction if [HgCl 2 ] = 0.094 M and [C 2 O 4 2– ] = 0.19 M?

(d) Are all four experiments necessary to answer parts (a) - (c)? Explain.

Use the method of initial rates to find the orders of reaction in each component. This will allow evaluation of the rate constant and the initial rate of reaction at any other condition.

(a) Rate = k [HgCl 2 ] m [C 2 O 4 2– ] n

Compare the rates in experiments 1 and 2 (or 3 and 4) to find the order in oxalate ion:

Compare the rates in experiments 2 and 3 (or 1 and 4) to find the order in mercury(II) chloride:

Therefore, the reaction is first order with respect to mercury(II) chloride and second order with respect to oxalate. The overall order is the sum of these, 2 + 1 = 3, third order.

(b) To find the rate constant, use the rate equation and solve for k :

Experiment [HgCl 2 ] (M) [C 2 O 4 2– ] (M) Initial rate (mol L –1 min –1 ) k

The average k = 7.6×10 –3 M –2 min –1

(c) Rate = k [HgCl 2 ][C 2 O 4 2– ] 2

Rate = (7.6×10 –3 M –2 min –1 )[0.094 M][0.19 M] 2 = 2.6×10 –5 M min –1

(d) Since there are only two reactants, three experiments are the minimum required to find the rate equation and rate constant. Experiments 1 - 3 would have sufficed to answer the questions posed.

10. In the reaction A → products, we find that when [A] has fallen to half of its initial value, the reaction proceeds at the same rate as its initial rate. Is the reaction zero order, first order, or second order? Explain.

Since the rate is not changing as a function of time, the reaction must be zero order.

11. In the reaction, A → products, with the initial concentration [A] 0 = 1.512 M, [A] is found to be 1.496 M at t = 30 s. With the initial concentration [A] 0 = 2.584 M, [A] is found to be 2.552 M at t = 1 min. What is the order of this reaction?

First, find the rate of reaction in each experiment, being sure to be consistent in units. Then use the Method of Initial rates to find the order of reaction.

Experiment [A] 0 (M) [A] (M) at t (s) Rate (M s –1 )

Since the rates are the same for both experiments, the reaction must be zero order. Mathematically:

12. A first order reaction, A → products, has a rate of reaction of 0.00250 M s –1 when [A] = 0.484 M. (a) What is the rate constant, k , for this reaction? (b) Does t 3/4 depend on the initial concentration? Does t 4/5 ? Explain.

(a) For a first order reaction, Rate = k [A]. Since the rate and concentration are known, solving the equation for k gives the required answer.

k = Rate/[A] = 0.00250 M s –1 /0.0484 M = 5.17×10 –3 s –1 .

(b) Neither t 3/4 nor t 4/5 depend upon intial concentration because this is a first order reaction, which means that the time associated reaction is simply related to the rate constant.

13. In the first-order decomposition of dinitrogen pentoxide at 335 K,

N 2 O 5 ( g ) → 2 NO 2 ( g ) + ½ O 2 ( g )

if we start with a 2.50-g sample of N 2 O 5 at 335 K and have 1.50 g remaining after 109 s, (a) What is the value of the rate constant k ? (b) What is the half-life of the reaction? (c) What mass of N 2 O 5 will remain after 5.0 min?

Since the concentration units cancel out, we can work directly in grams for this problem.

(a) [N 2 O 5 ] 0 = 2.50 g, [N 2 O 5 ] t = 1.50 g, t = 109 s, so

so k = 4.69×10 –3 s –1

(b) t ½ = 0.693/ k = 0.693/4.69×10 –3 = 148 s.

(c) t = 5.0 min = 300. s, so

14. The smog constituent peroxyacetyl nitrate (PAN) dissociates into peroxyacetyl radicals and NO 2 ( g ) in a first order reaction with a half-life of 32 min.

If the initial concentration of PAN in an air sample is 2.7×10 15 molecules/L, what will be the concentration 2.24 h later?

Thus, k = 0.693/t ½ = 0.693/32 = 0.022 min –1 . For an initial concentration [PAN] 0 = 2.7 ×10 15 molecules/L and at t = 2.24 h = 134 min:

[PAN] = [PAN] o e – k t = (2.7×10 15 )e –(0.022)(134) = 1.4×10 14 molecules/L.

15. The following data were obtained in two separate experiments in the reaction: A → products. Determine the rate law for this reaction, including the value of k .

Experiment 1 Experiment 2

[A] (M) time (s) [A] (M) time (s)

0.800 0 0.400 0

0.775 40 0.390 64

0.750 83 0.380 132

0.725 129 0.370 203

0.700 179 0.360 278

The general rate law is Rate = – k [A] m ; the method of initial rates can be used to establish the order of reaction and give an estimate of the rate constant.

The initial rate are found from the first two data points in each experiment:

Using the initial concentrations with the initial rates:

So this is a second order reaction. The rate constant can estimated from k = –Rate/[A] 2 for each experiment: k (exp. 1) = –(–6.25×10 –4 )/[0.800] 2 = 9.8×10 –4 M –1 s –1 and k (exp. 2) = –(–1.56×10 –4 )/[0.400] 2 = 9.8×10 –4 M –1 s –1 .

k = 9.8×10 –4 M –1 s –1

To confirm this and get a better value for the rate constant, a plot of [A] –1 vs. t should be linear with a slope equal to the rate constant:

The slopes of these plots show that a better value for the rate constant is k = 1.0×10 –3 M –1 s –1 .

16. Listed below are initial rates, expressed in terms of the rate of decrease of partial pressure of a reactant for the following reaction at 826 °C. Determine the rate law for this reaction, including the value for k .

NO( g ) + H 2 ( g ) → ½N 2 ( g ) + H 2 O( g )

With initial P H2 = 400 mmHg With initial P NO = 400 mmHg

Initial P NO (mmHg) Rate (mmHg/s) Initial P H2 (mmHg) Rate (mmHg/s)

359 0.750 289 0.800

300 0.515 205 0.550

152 0.125 147 0.395

The unknown rate law is given by Rate = k [NO] m [H 2 ] n . Using the Method of Initial rates will give the rate law and the value of the rate constant. Since the units cancel in the Method of Initial rates, we do not need to convert to molarity.

To find the order in NO, use the first set of data where the pressure of H 2 is kept constant. Experiments 2 and 3 are about a factor of 2 difference in initial pressure of NO, so will give easier numbers to work with:

Within experimental error, m = 2, or second order in NO.

To find the order in H 2 , use experiments 1 and 3 for the easiest numbers:

Within experimental error, n = 1, or first order in H 2 .

The rate constant can now be found from the rate law: k = Rate/[NO] 2 [H 2 ] for each experiment:

Initial P H2 (mmHg) Initial P NO (mmHg) Initial Rate (mmHg/s) k (mmHg –2 s –1 )

400 359 0.750 (0.750)/(400)(359) 2 = 1.45×10 –8

400 300 0.515 (0.515)/(400)(300) 2 = 1.43×10 –8

400 152 0.125 (0.125)/(400)(152) 2 = 1.35×10 –8

289 400 0.800 (0.800)/(289)(400) 2 = 1.73×10 –8

205 400 0.550 (0.550)/(205)(400) 2 = 1.68×10 –8

147 400 0.395 (0.395)/(147)(400) 2 = 1.68×10 –8

Averaging all of the k values gives k = 1.55×10 –8 mmHg –2 s –1

17. Rate constants for the first-order decomposition of acetonedicarboxylic acid

CO(CH 2 COOH) 2 ( aq ) → CO(CH 3 ) 2 ( aq ) + 2 CO 2 ( g )

acetonedicarboxylic acid acetone

are k = 4.75 ×10 –4 s –1 at 293 K and k = 1.63 ×10 –3 at 303 K. What is the activation energy, E a , for this reaction?

Use the two-point form of the Arrhenius equation to answer this question:

where k 2 = 1.63 ×10 –3 at T 2 = 303 K and k 1 = 4.75 ×10 –4 at T 1 = 293 K.

18. The following is proposed as a plausible reaction mechanism:

A + B → I (slow)

I + B → C + D (fast)

What is (a) the net reaction described by this mechanism and (b) a plausible rate law for the reaction?

(a) To find the net reaction, sum up all the reactions in the mechanism and eliminate common species on each side of the equation:

A + B + I + B → I + C + D

net: A + 2B → C + D

(b) The rate law is determined by the slow step in the reaction mechanism. Since a reaction mechanism is composed of elementary reactions, the rate law for the slow step can be written by inspection, with the orders of reaction being the stoichiometric coefficients:

Rate = k[A][B]

19. The following reaction exhibits the rate law: Rate = k [NO] 2 [Cl 2 ].

2 NO( g ) + Cl 2 ( g ) → 2 NOCl( g )

Explain why the following mechanism is not plausible for this reaction.

NO( g ) + Cl 2 → ← NOCl( g ) + Cl( g ) fast

NO( g ) + Cl( g ) → NOCl( g ) slow

The rate law is determined by the slow step: Rate = k slow [NO][Cl]

However, this rate law includes and intermediate that cannot be part of the rate law for the overall reaction - only NO or Cl 2 can be part of the rate law. The [Cl] term can be eliminated using the fast, reversible step where the rate of the forward reaction is the same as the rate of the reverse reaction:

Rate forward = Rate reverse

Rate forward = k forward [NO][Cl 2 ]

Rate reverse = k reverse [NOCl][Cl]

so k forward [NO][Cl 2 ] = k reverse [NOCl][Cl]

substituting for [Cl] in the slow step rate law gives:

This rate does not match the experimental rate law so this cannot be a plausible mechanism.

20. Show that the proposed mechanism is consistent with the rate law for the following reaction in aqueous solution,

Hg 2 2+ ( aq ) + Tl 3+ ( aq ) → 2 Hg 2+ ( aq ) + Tl + ( aq )

for which the observed rate law is

Proposed Mechanism:

Hg 2 2+ ( aq ) → ← Hg 2+ ( aq ) + Hg( s ) fast

Hg( s ) + Tl 3+ ( aq ) → Hg 2+ ( aq ) + Tl + ( aq ) slow

First, check to see that the proposed mechanism matches the experimental stoichiometry.

Add the two reactions together to give:

Hg 2 2+ ( aq ) + Hg( s ) + Tl 3+ ( aq ) → Hg 2+ ( aq ) + Hg( s ) + Hg 2+ ( aq ) + Tl + ( aq )

The Hg( s ) common to both sides of the equation cancel out and the two moles of Hg 2+ ( aq ) can be collected to give the observed reaction, so the stoichiometry matches.

The proposed mechanism can be rewritten as

Hg 2 2+ ( aq ) → k 1 Hg 2+ ( aq ) + Hg( s ) Rate = k 1 [Hg 2 2+ ] (fast)

Hg 2+ ( aq ) + Hg( s ) → k –1 Hg 2 2+ ( aq ) Rate = k –1 [Hg 2+ ][Hg] (fast)

Hg( s ) + Tl 3+ ( aq ) → k 2 Hg 2+ ( aq ) + Tl + ( aq ) Rate = k 2 [Hg 2+ ][Hg] (slow)

The overall rate is determined by the slow step, but this has an intermediate ([Hg}) that must be eliminated in order to evaluate the mechanism. This can be done using the two fast steps, which must have approximately the same rate, or:

k 1 [Hg 2 2+ ] = k –1 [Hg 2+ ][Hg]

[Hg] = k 1 [Hg 2 2+ ]/ k –1 [Hg 2+ ]

Using this expression in the rate law for the slow step gives

Rate = k 2 [Hg 2+ ] k 1 [Hg 2 2+ ]/ k –1 [Hg 2+ ]

Rate = ( k 2 k 1 / k –1 )[Hg 2+ [Hg 2 2+ ]/[Hg 2+ ]

which exactly matches the observed rate law when k = k 2 k 1 / k –1

21. Benzenediazonium chloride decomposes in water yielding N 2 ( g ).

C 6 H 5 N 2 Cl( aq ) → C 6 H 5 Cl( aq ) + N 2 ( g )

The data tabulated below were obtained for the decomposition of a 0.071 M solution at 50 °C ( t = ∞ corresponds to the completed reaction). To obtain [C 6 H 5 N 2 Cl] as a function of time, note that during the first 3 min, the volume of N 2 ( g ) produced was 10.8 mL of a total of 58.3 mL, corresponding to this fraction of the total reaction: 10.8 mL/58.3 mL = 0.185. An equal fraction of the available C 6 H 5 N 2 Cl was consumed during the same time.

time (min) N 2 ( g ) (mL)

∞ 58.3

(a) Plot graphs showing the disappearance of C 6 H 5 N 2 Cl and the formation of N 2 ( g ) as a function of time.

(b) What is the initial rate of formation of N 2 ( g )?

(c) What is the rate of disappearance of C 6 H 5 N 2 Cl at t = 20 min?

(d) What is the half-life, t ½ , of the reaction?

(e) Write the rate law for this reaction, including a value for k .

(a) First, find the concentrations of C 6 H 5 N 2 Cl remaining at each time:

time (min) N 2 ( g ) (mL) fraction N 2 = mL/58.3 mL C 6 H 5 N 2 Cl (M) = 0.071×(1 – fraction)

0 0 0/58.3 = 0.00 0.071(1 – 0) = 0.071

3 10.8 10.8/58.3 = 0.185 0.071(1 – 0.185) = 0.058

6 19.3 19.3/58.3 = 0.331 0.071(1 – 0.331) = 0.047

9 26.3 26.3/58.3 = 0.451 0.071(1 – 0.451) = 0.039

12 32.4 32.4/58.3 = 0.556 0.071(1 – 0.556) = 0.032

15 37.3 37.3/58.3 = 0.640 0.071(1 – 0.640) = 0.026

18 41.3 41.3/58.3 = 0.708 0.071(1 – 0.708) = 0.021

21 44.3 44.3/58.3 = 0.760 0.071(1 – 0.760) = 0.017

24 46.5 46.5/58.3 = 0.798 0.071(1 – 0.798) = 0.014

27 48.4 48.4/58.3 = 0.830 0.071(1 – 0.830) = 0.012

30 50.4 50.4/58.3 = 0.864 0.071(1 – 0.864) = 0.0097

∞ 58.3 58.3/58.3 = 1.00 0.071(1 – 1.00) = 0.00

The plots are:

(b) From the definition of rate,

(c) Again, using the definition of rate and approximating t = 20 min using the t = 18 min and t = 21 min points,

(d) The initial concentration of benzenediazonium chloride is 0.071 M, so the first half-life is reached when the concentration has reached 0.0355 M. Using the graph from part ( a ) gives t ½ ~ 10 min.

(e) The order of the reaction can be determined by looking at the time of the second half-life, when the benzenediazonium chloride concentration has been reduced to ½ of its initial value, 0.018 M. This occurs at ~21 min, just twice the time for the first half-life. Since t ½ is the same for the two different time intervals, this means that the reaction is first order.

Rate = k [C 6 H 5 N 2 Cl]

Plotting ln[C 6 H 5 N 2 Cl] vs. t confirms this and the slope of the plot gives the rate constant, k .

The slope of the plot = –0.0664, so k = 0.0664 min –1 . (This allows a better estimate of the half-life to be t ½ = 0.693/ k = 0.693/0.0664 = 10.4 min.)

22. Hydroxide ion is involved in the mechanism but not consumed in this reaction in aqueous solution.

OCl – ( aq ) + I – ( aq ) → OH – OI – ( aq ) + Cl – ( aq )

(a) From the data in the table, determine the order of reaction with respect to OCl – , I – , and OH – , and the overall order.

[OCl – ] (M) [I – ] (M) [OH – ] (M) Rate of formation of OI – (mol L –1 s –1 )

0.0040 0.0020 1.00 4.8×10 –4

0.0020 0.0040 1.00 5.0×10 –4

0.0020 0.0020 1.00 2.4×10 –4

0.0020 0.0020 0.50 4.6×10 –4

0.0020 0.0020 0.25 9.4×10 –4

(b) Write the rate law, and determine the value of the rate constant, k .

(c) Show that the following mechanism is consistent with the net equation and with the rate law. Which is the rate–determining step?

OCl – ( aq ) + H 2 O( l ) → ← HOCl( aq ) + OH – ( aq ) I – ( aq ) + HOCl( aq ) → HOI( aq ) + Cl – ( aq ) HOI( aq ) + OH – ( aq ) → H 2 O( l ) + OI – ( aq )

(d) Is it appropriate to refer to OH – as a catalyst in this reaction? Explain.

(a) Use the method of initial rates for a rate law of the general form Rate = k [OCl – ] m [I – ] n [OH – ] p . Experiments 1 and 3 give the order in hypochlorite ion (iodide and hydroxide are constant): doubling the concentration of hypochlorite doubles the initial rate, meaning that the reaction is 1 st order in hypochlorite ion, m = 1. Experiments 2 and 3 can be used to find the order in iodide ion (hypochlorite and hydroxide are unchanged): doubling the iodide concentration doubles the rate so the reaction is also first order in iodide ion, n = 1. Experiments 3 and 4 can be used to find the order in hydroxide (hypochlorite and iodide are unchanged): doubling the concentration halves the rate, so the order in hydroxide must be –1, p = –1. Mathematically:

The overall order is the sum of each individual order = 1 + 1 + –1 = 1

(b) Based on the orders found above, the rate law is:

The rate constant can be found by using the rate law and the data:

[OCl – ] (M) [I – ] (M) [OH – ] (M) Rate of formation of OI – (mol L –1 s –1 ) 0.0040 0.0020 1.00 4.8×10 –4 60.0 s –1 0.0020 0.0040 1.00 5.0×10 –4 62.5 s –1 0.0020 0.0020 1.00 2.4×10 –4 60.0 s –1 0.0020 0.0020 0.50 4.6×10 –4 57.5 s –1 0.0020 0.0020 0.25 9.4×10 –4 58.75 s –1

The average rate constant (to the correct number of significant figures) = 60. s –1

OCl – ( aq ) + H 2 O( l ) + I – ( aq ) + HOCl( aq ) + HOI( aq ) + OH – ( aq ) → HOCl( aq ) + OH – ( aq ) + HOI( aq ) + Cl – ( aq ) + H 2 O( l ) + OI – ( aq )

After eliminating common species:

OCl – ( aq ) + I – ( aq ) → OI – ( aq ) + Cl – ( aq )

which is the correct stoichiometry.

To find the rate–determining step, consider the rate law for each elementary reaction:

Step 1, forward reaction: Rate = k forward [OCl – ][H 2 O]

Step 1, reverse reaction: Rate = k reverse [HOCl][ OH – ]

Step 2: Rate = k 2 [I – ][HOCl]

Step 3: Rate = k 3 [HOI][OH – ]

Step 1, forward does not have enough terms in the rate law. Step 1, reverse and Step 3 both have hydroxide in the numerator but the experimental rate law requires it to be in the denominator. This leaves Step 2 as the likely slow step, but this has an intermediate that must be eliminated by using Step 1, assuming it to be a fast step:

For Step 1, Rate forward = Rate reverse so k forward [OCl – ][H 2 O] = k reverse [HOCl][OH – ]

Solving for the intermediate, [HOCl] gives:

Substituting this into the rate law for Step 2, the proposed slow step, gives:

Since water is the solvent, its concentration does not change and should be grouped with the constants:

This now matches the experimental rate law.

(d) Since hydroxide is found in the denominator of the rate law, an increase in concentration slows the reaction down so OH – is an inhibitor, not a catalyst. Compare experiments 3, 4, and 5 to see how decreasing the hydroxide increases the reaction rate.

## Chemistry - Chemical Kinetics: Solved Example Problems | 12th Chemistry : UNIT 7 : Chemical Kinetics

Chapter: 12th chemistry : unit 7 : chemical kinetics, chemical kinetics: solved example problems, molecularity: solved example problems.

Consider the oxidation of nitric oxide to form NO 2

2NO(g) + O 2 (g) → 2NO 2 (g)

(a). Express the rate of the reaction in terms of changes in the concentration of NO,O 2 and NO 2 .

(b). At a particular instant, when [O 2 ] is decreasing at 0.2 mol L −1 s −1 at what rate is [NO2 ] increasing at that instant?

Evaluate yourself 1

1) Write the rate expression for the following reactions, assuming them as elementary reactions.

i) 3A + 5B 2 →4CD

ii) X 2 + Y 2 →2XY

2). Consider the decomposition of N 2 O 5 (g) to form NO 2 (g) and O 2 (g) . At a particular instant N 2 O 5 disappears at a rate of 2.5x10 -2 mol dm -3 s -1 . At what rates are NO 2 and O 2 formed? What is the rate of the reaction?

1. What is the order with respect to each of the reactant and overall order of the following reactions?

(a). 5Br − ( aq ) + BrO 3 − (aq ) + 6H + (aq)

→ 3Br 2 (l ) + 3H 2 O(l)

The experimental rate law is

Rate = k [Br − ][BrO 3 ][H + ] 2

(b). CH 3 CHO(g ) → Δ → CH 4 (g) + CO(g)

the experimental rate law is

Rate = k [CH 3 CHO] 3/2

a) First order with respect to Br − , first order with respect to BrO 3 − and second order with respect to H+ . Hence the overall order of the reaction is equal to 1 + 1 + 2 = 4

b) Order of the reaction with respect to acetaldehyde is 3/2 and overall order is also 3/2

2. The rate of the reaction x + 2y → product is 4 x 10 −3 mol L −1 s −1 if [x]=[y]=0.2 M and rate constant at 400K is 2 x 10 -2 s -1 , What is the overall order of the reaction.

Rate = k [x] n [y] m

4 x 10 -3 mol L -1 s -1 = 2 x 10 -2 s -1 (0.2 mol L -1 ) n (0.2mol L -1 ) m

Comparing the powers on both sides

The overall order of the reaction n + m = 1

Evaluate yourself 2

1). For a reaction, X + Y → product ; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall order of the reaction?

2). Find the individual and overall order of the following reaction using the given data.

2NO(g) + Cl 2 (g) → 2NOCl(g)

## Half life period of a reaction: Solved Example Problems

A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989 ; log10 = 1)

For a first order reaction,

Let [A 0 ] = 100M

t = t 90% ; [A]=10M (given that t 90 % =8hours)

t = t 80% ; [A]=20M

Find the value of k using the given data

Substitute the value of k in equation (2)

t 80% = 8hours x 0.6989

t 80% = 5.59hours

(ii) The half life of a first order reaction x → products is 6.932 x 10 4 s at 500K . What percentage of x would be decomposed on heating at 500K for 100 min. (e 0.06 = 1.06)

Given t 1/2 = 0.6932 x 10 4 s

To solve :2 when t=100 min,

[ [A 0 ] −[A] / [A 0 ] ] x 100 = ?

We know that

For a first order reaction, t 1/2 = 0.6932 / k

k = 10 −5 s −1

Show that in case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.

Evaluate yourself:

1. In a first order reaction A → products 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction?

2. The rate constant for a first order reaction is 2.3 X 10 −4 s −1 If the initial concentration of the reactant is 0.01M . What concentration will remain after 1 hour?

3. Hydrolysis of an ester in an aqueous solution was studied by titrating the liberated carboxylic acid against sodium hydroxide solution. The concentrations of the ester at different time intervals are given below.

Show that, the reaction follows first order kinetics.

## Arrhenius equation - The effect of temperature on reaction rate: Solved Example Problems

The rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s-1 respectively. Calculate the value of activation energy.

According to Arrhenius equation

Ea = 2305 J mol−1

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation

Where Ea is the activation energy. When a graph is plotted for log k Vs 1/T a straight line with a slope of - 4000K is obtained. Calculate the activation energy

E a = − 2.303 R m

E a = − 2.303 x 8.314 J K −1 mol −1 x (− 4000K )

E a = 76,589J mol −1

E a = 76.589 kJ mol −1

## Evaluate yourself

For a first order reaction the rate constant at 500K is 8 X 10 −4 s −1 . Calculate the frequency factor, if the energy of activation for the reaction is 190 kJ mol -1 .

## Chemical Kinetics - Example : Solved Example Problems

1. The rate law for a reaction of A, B and C hasbeenfoundtobe rate = k [ A ] 2 [ B ][ L ] 3/2

How would the rate of reaction change when

(i) Concentration of [L] is quadrupled

(ii) Concentration of both [A] and [B] are doubled

(iii) Concentration of [A] is halved

(iv) Concentration of [A] is reduced to(1/3) and concentration of [L] is quadrupled.

2. The rate of formation of a dimer in a second order reaction is 7.5 × 10 − 3 mol L − 1 s − 1 at 0.05 mol L − 1 monomer concentration. Calculate the rate constant.

Let us consider the dimerisation of a monomer M

Rate= k [M] n

Given that n=2 and [M] = 0.05 mol L -1

Rate = 7.5 X 10 -3 mol L -1 s -1

k = Rate/[M] n

k = 7.5x10 -3 /(0.05) 2 = 3mol -1 Ls -1

3. For a reaction x + y + z → products the rate law is given by rate = k [ x ] 3/2 [ y ] 1/2 what is the overall order of the reaction and what is the order of the reaction with respect to z.

i.e., second order reaction.

Since the rate expression does not contain the concentration of z , the reaction is zero order with respect to z.

15. The decomposition of Cl 2 O 7 at 500K in the gas phase to Cl 2 and O 2 is a first order reaction. After 1 minute at 500K, the pressure of Cl 2 O 7 falls from 0.08 to 0.04 atm. Calculate the rate constant in s -1 .

4. A gas phase reaction has energy of activation 200 kJ mol -1 . If the frequency factor of the reaction is 1.6 × 10 13 s − 1 Calculate the rate constant at 600 K. ( e − 40 .09 = 3.8 × 10 − 18 )

20. For the reaction 2 x + y → L find the rate law from the following data.

5. The rate constant for a first order reaction is 1.54 × 10 -3 s -1 . Calculate its half life time.

We know that, t 1/2 = 0.693/ k

t 1/2 = 0.693/1.54 x 10 -3 s -1 = 450 s

6. The half life of the homogeneous gaseous reaction SO 2 Cl 2 → SO 2 + Cl 2 which obeys first order kinetics is 8.0 minutes. How long will it take for the concentration of SO 2 Cl 2 to be reduced to 1% of the initial value?

We know that, k = 0.693/ t 1/2

k = 0.693/ 8.0 minutes= 0.087 minutes -1

7. The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?

8. A zero order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?

i) Let A = 100M, [A 0 ]–[A] = 20M,

For the zero order reaction

k=([A 0 ]-[A] / t)

k=(20M / 20min) = 1 Mmin -1

Rate constant for a reaction = 1Mmin -1

ii) To calculate the time for 80% of completion

k = 1Mmin -1 , [A 0 ] = 100M, [A 0 ]-[A] = 80M, t = ?

t=([A 0 ]-[A] / k) = (80M / 1Mmin -1 ) = 80min

9. The activation energy of a reaction is 225 k Cal mol -1 and the value of rate constant at 40°C is 1.8 × 10 − 5 s − 1 . Calculate the frequency factor, A.

Here, we are given that

E a = 22.5 kcal mol -1 = 22500 cal mol -1

T = 40 ° C = 40 + 273 = 313 K

k = 1.8 × 10 -5 sec -1

Substituting the values in the equation

log A = log (1.8) −5 + (15.7089)

log A = 10.9642

A = antilog (10.9642)

A = 9.208 × 10 10 collisions s −1

10. Benzene diazonium chloride in aqueous solution decomposes according to the equation C 6 H 5 N 2 Cl → C 6 H 5 Cl + N 2 . Starting with an initial concentration of 10 g L − 1 , the volume of N 2 gas obtained at 50 °C at different intervals of time was found to be as under:

Show that the above reaction follows the first order kinetics. What is the value of the rate constant?

For a first order reaction

In the present case, V∞ = 58.3 ml.

The value of k at different time can be calculated as follows:

Since the value of k comes out to be nearly constant, the given reaction is of the first order. The mean value of k = 0.0676 min -1

11. From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order:

Where t is the time in minutes and V is the volume of standard KMnO4 solution required for titrating the same volume of the reaction mixture.

In the present case, V o = 46.1 ml.

The value of k at each instant can be calculated as follows:

Thus, the value of k comes out to be nearly constant. Hence it is a reaction of the first order.

12. Where t is the time in minutes and V is the volume of standard KMnO 4 solution required for titrating the same volume of the reaction mixture.

i) For the first order reaction k = 2.303/t log [A 0 ]/[A]

Assume, [A 0 ] = 100 %, t = 50 minutes

Therefore, [A] = 100 – 40 = 60

k = (2.303 / 50) log (100 / 60)

k = 0.010216 min -1

Hence the value of the rate constant is 0.010216 min -1

ii) t = ?, when the reaction is 80% completed,

[A] = 100 – 80 = 20%

From above, k = 0.010216 min -1

t = (2.303 / 0.010216) log (100 / 20)

t = 157.58 min

The time at which the reaction will be 80% complete is 157.58 min.

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- Chemistry Concept Questions and Answers

## Chemical Kinetics Questions

Modern chemical (reaction) kinetics is a branch of science that studies and describes chemical reactions as we know them now. It is the study of the rate of chemical reactions or the transformations of reactants into products that take place according to a specific mechanism, i.e. the reaction mechanism.

Kinetic studies are useful for comprehending reactions and have practical applications. In industry, for example, reactions are carried out in reactors, where compounds are mixed together, perhaps heated and stirred for a short time, and then transferred on to the next stage of the process. It’s critical to know how long to hold a reaction at one stage before moving on to the next, in order to ensure that the previous reaction has been completed before moving on to the next.

## Chemical Kinetics Chemistry Questions with Solutions

Q1: Define Reaction Rate.

The reaction rate is determined by how rapidly the products are created and the reactants are consumed. It is usual to deal with substance concentrations in chemical systems, which are defined as the amount of substance per unit volume. The concentration of a material consumed or produced in unit time can thus be described as the rate. It’s sometimes easier to express rates as the number of molecules generated or consumed per unit of time.

Q2: What is meant by the Half-Life?

The half-life of a reactant, which is defined as the time it takes for half of the initial amount to undergo reaction, is a useful rate metric. The half-life is independent of the initial amount for a certain form of kinetic behaviour (first-order kinetics).

Radioactive substances are a simple and typical example of a half-life that is independent of the initial amount. Uranium-238, for example, has a half-life of 4.5 billion years, meaning that half of an initial amount of uranium will have decayed in that time. Many chemical reactions exhibit the same behaviour.

Q3: Write the rate equation for the reaction 2A + B → C if the order of the reaction is zero.

The rate equation for the reaction 2A + B → C in the zero order reaction is,

Rate = k [A] 0 [B] 0 = k.

Q4: Although the reaction between H 2 (g) and O 2 (g) is highly feasible, leaving the gases in the same vessel at room temperature does not result in the creation of water. Explain.

Although the reaction between H 2 (g) and O 2 (g) is very feasible, leaving the gases to remain at room temperature in the same vessel does not result in the creation of water because the reactants’ activation energy is very high at room temperature and not readily available.

Q5: Match the statements given in Column I and Column II

a) – (iii); (b) – (i); (c) – (iv); (d) – (vi); (e) – (ii); (f) – (v)

Q6: Match the items of Column I and Column II.

(a) – (ii); (b) – (i); (c) – (iv); (d) – (iii)

Q7: Define each of the following:

i) Specific rate of a reaction

ii) Activation energy of a reaction

i) Specific rate of a reaction : When the molar concentration of each of the reactants is equal, the rate of reaction is called the specific rate of reaction.

ii) Activation energy of a reaction : The activation energy is the minimum extra amount of energy absorbed by the reactant molecules in need of their energy to equal the threshold value. The addition of a catalyst reduces the reaction’s activation energy.

Q8: Which among the following is a false statement?

(i) Rate of zero order reaction is independent of initial concentration of reactant.

(ii) Half-life of a third order reaction is inversely proportional to the square of initial concentration of the reactant.

(iii) Molecularity of a reaction may be zero or fraction.

(iv) For a first order reaction, t 1/2 = 0.693/K

Answer: (iii) Molecularity of a reaction may be zero or fraction.

Q9: The average rate and instantaneous rate of a reaction are equal

(i) at the start

(ii) at the end

(iii) in the middle

(iv) when two rates have a time interval equal to zero

Answer: (iv) when two rates have a time interval equal to zero

Q10: For a chemical reaction A→B, it is found that the rate of reaction doubles when the concentration of A is increased four times. The order of the reaction is

Answer: (iv) Half

Q11: Differentiate between the Order and Molecularity of a reaction.

Q12: With the help of an example explain what is meant by pseudo first order reaction.

Consider the reaction of ethyl acetate hydrolysis, which can be expressed as

CH 3 COOC 2 H 5 + H 2 O → CH 3 COOH + C 2 H 5 OH

The rate equation can be given as Rate = k [CH 3 COOC 2 H 5 ] [H 2 O]

The concentration of water is relatively high in this case and consequently does not change much during the process. As a result, the rate of the reaction is independent of the change in H 2 O concentration. We might write the effective rate equation as if the term for a change in water concentration in the above reaction is set to zero.

Rate = k [CH 3 COOC 2 H 5 ]

The term here takes into account the value of the water’s constant concentration.

where K = K’ [H 2 O]

The reaction is a first-order reaction, as we can see. These reactions are termed pseudo-first-order reactions .

Q13: What is the use of the integrated rate equation?

Uses of integrated rate equation are as follows:

(i) When the concentration of a reactant at different intervals is known, the value of the rate constant can be calculated.

(ii) The knowledge of reaction concentration at different intervals can be used to establish the order of a reaction.

Q14: How does collision theory explain the formation of products in a chemical reaction?

The reactant molecules are considered to be hard spheres in Collision Theory, and the reaction occurs when these molecules collide. The production of a product is induced by collisions between molecules with sufficient kinetic energy (called threshold energy) and proper orientation. The conditions for effective collision, and hence the rate of a reaction, are determined by the activation energy and proper orientation of the molecules.

Q15: The time required to decompose SO 2 Cl 2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

We know that for a 1st order reaction,

It is already given that t 1/2 = 60mins

= 0.693/60 = 0.01155 min -1 = 1.155 min -1

## Practise Questions on Chemical Kinetics

Q1: Mention the factors that affect the rate of a chemical reaction.

Q2: For a reaction, A + B → Product; the rate law is given by, R = k [A] 1/2 [B] 2 . What is the order of the reaction?

Q3: Define the following terms:

i) Pseudo first order reaction.

ii) Half-life period of reaction (t 1/2 ).

Q4: Distinguish between Rate Expression and Rate Constant of a reaction.

Q5: Write the Arrhenius equation.

Click the PDF to check the answers for Practice Questions. Download PDF

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## Solved Examples

Example : 1.

The experimental data for the reaction 2A + B2 → 2AB is as follows:

From an examination of above data, it is clear that when the concentration of B 2 is doubled, the rate is doubled. Hence the order of reaction with respect to B 2 is one. Further when concentration of A is doubled, the rate remain unaltered. So, order of reaction with respect to A is zero. The probable rate law for the reaction will be -dx/dt = k[B 2 ][A] 0 = k[B 2 ] Alternatively Rate = k[B 2 ] α 1.6 × 10-4 = k[0.5] α 3.2 × 10-4 = k[1] α On dividing we get α = 1 .·. Rate = k[A] 0 [B 2 ] = k[B 2 ]

__________________________________________________________________________

## Example : 2

For the reaction A + 2B → 2C the following data were obtained.

Write down the rate law for the reaction.

Let the rate law be -dx/dt = k[A] x [B] y By keeping the concentration of B constant in experiments (1), (2) and (3) and increasing concentration uniformly, the rate also increases uniformly, thus, Rate ∞ [A], i.e., x = 1 By keeping the concentration of A constant in experiments (1), (4) and (5) and increasing the concentration of B, the rate remains the same. Hence, y = 0 The rate law is -dx/dt = k[A]

Alternatively method: From Expt. (1), k[1.0] x [1.0] y = 0.15 ....(i) From Expt. (2), k[2.0] x [1.0] y = 0.30 ... (ii) Dividing Eq. (ii) by Eq. (i), [2.0] x /[1.0] x = 0.30/0.15 = 2

So , x = 1

From Expt. (1), k[1.0] x [1.0] y = 0.15 ....(i) From Expt. (4), k[1.0] x [2.0] y = 0.15 .... (iii) Dividing Eq. (iii) by Eq. (i), [2.0] y /[1.0] y = 1 So , y = 0 Hence, the rate law is -dx/dt = k[A].

___________________________________________________________________________________

## Example : 3

For the reaction: 2NO + Cl2 → 2NOCl at 300 K following data are obtained.

Write rate of law for the reaction. What is the order of the reaction? Also calculate the specific rate constant. .

Let the rate law for the reaction be

Rate = k[NO] x [Cl 2 ] y

From Expt. (1),

1.2×10-4 = k[0.010] x [0.010] y ... (i)

From Expt. (2),

2.4×10 -4 = k[0.010] x [0.020] y ... (ii)

Dividing Eq. (ii) by Eq. (i),

(9.6×10-4)/(2.4×10-4) = ([0.020] x )/[0.010] x or 2 = (2) y y = 1 From Expt. (2), 2.4×10-4 = k[0.010] x [0.020] y .... (ii) From Expt. (3), 9.6×10-4 = k[0.020] x [0.020] y .... (ii) Dividing Eq. (ii) by Eq. (ii), (9.6 × 10-4)/(2.4 × 10-4) = ([0.020]x )/[0.010]x or 4 = 2x x = 2 Order of reaction = x + y = 2 + 1 = 3 Rate law for the reaction is Rate = k[NO] 2 [Cl 2 ] Considering Eq. (i) again, 1.2 × 10 -4 = k[0.010] 2 [0.010] k = (1.2×10-4)/[0.010] 3 = 1.2×10 2 mol -2 litre 2 sec-1 ________________________________________________________________________________________________________

For the hypothetical reaction 2A + B → products the following data are obtained.

Find out how the rate of the ration depends upon the concentration of A and B and fill in the blanks.

## Solution :

From Expt. (2) and (3), it is clear that when concentration of A is kept constant and that of B is doubled, the rate increases four times. This shows that the reaction is of second order with respect to B. Similarly, from Expt. (1) and (2), it is observed that when concentration of A is increased three times and that of B two times, the rate becomes twelve times. Hence, the reaction is first order with respect to A. Thus the rate law for the reaction is Rate = k[A][B]2

From Expt. (1), 1.75×10-4 = k[0.0017] x [0.0017] y [1.0] z From Expt. (4), 3.50×10-4 = k[0.0017] x [0.0017] y [0.5] z (1.75 × 10-4)/(3.50 × 10-4) = [1.0] z /[0.5] z or 1/2 = 2z or 2-1 = 2z z = -1, i.e., order w.r.to OH - is -1. Rate law = k[OCI- ][I- ]/[OH-] From Expt.(1)k = (1.75×10 -4 [OH - ])/[OCI - ][I - ] = (1.75 × 10 -4 ×1.0)/(0.0017×0.0017) = 60.55 s -1 ____________________________________________________________________________

## Example : 5

The rate law for the reaction, 2Cl 2 O → 2Cl 2 + O 2 at 200oC is found to be : rate = k[Cl 2 O] 2 (a) How would the rate change if [Cl 2 O] is reduced to one-third of its original value? (b) How should the [Cl 2 O] be changed in order to double the rate? (c) How would the rate change if [Cl 2 O] is raised to threefold of its original value?

## Solution :

(a) Rate equation for the reaction, r = k[Cl 2 O] 2 Let the new rate be r'; so r' = k[(Cl 2 O)/3] 2 = 1/9 r (b) In order to have the rate = 2r, let the concentration of Cl 2 O be x. So 2r = kx 2 .... (i) We know that r = k[Cl 2 O] 2 .... (ii) Dividing Eq. (i) by (ii), 2r/r = (kx 2 )/(k[Cl 2 O] 2 ) or 2 = x 2 /[Cl 2 O] 2 or x 2 = 2[Cl 2 O] 2 or x = √√2 [Cl 2 O] (d) New rate = k[3Cl 2 O] 2 = 9k[Cl 2 O] 2 = 9r _____________________________________________________________________________________________________

## Example : 6

For a reaction in which A and B from C, the following data were obtained from three experiments:

What is the rate equation of the reaction and what is the value of rate constant?

## Solution:

Let the rate equation be k[A] x [B] y From Expt. (1), 0.3×10 -4 = k[0.03] x [0.03] y ... (i) From Expt. (2), 1.2×10 -4 = k[0.06] x [0.06] y ... (ii) (1.2 × 10-4)/(0.3×10-4) = ([0.06] x [0.06] y )/([0.03] x [0.03] y ) = 2 x × 2y = 4 ... (iii) Similarly from Expt. (1) and (3), 2x ×3y = 9 ....(iv) Solving Eq. (iii) and (iv), x = 0, y = 2 Rate equation, Rate = k[B] 2 Considering Eq. (i) again, k = (0.3×10 -4 )/[0.03] 2 = 3.33 × 10-2 mol L -1 s -1 ________________________________________________________________________________

## Example : 7

For a first order reaction when log k was plotted against , a straight 1/T line with a slope of -6000 was obtained. Calculate the activation energy of the reaction

Slope = -Ea/2.303R = –6000

= 27483.935Cal = 27.48 Kcal

## Example : 8

Solution: .

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## Example : 9

The time required for 10% completion of first order reaction at 298 K is equal to that required for its 25% completion at 308K. If the preexponential factor for the reaction is 3.56 ´ 10 9 s –1 , calculate the energy of activation

_________________________________________________________________________________

## Question : 10

At 380°C, the half-life period for the first order decomposition of H 2 O 2 is 360 min. the energy of activation of the reaction is 200 kJ mol –1 , Calculate the time required for 75% decomposition at 450°C.

## Related Resources:-

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You can also refer to Collision Theory of Reaction Rate

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