# Kinetics by samsond

VIEWS: 91 PAGES: 15

• pg 1
```									1. Consider the reaction,
2NO(g) + 2H2(g) → N2(g) + 2H2O(g).
Using the data given below,
a) determine the rate expression and
b) find the value of rate constant.

[NO]      [H2]   Initial rate
M         M      M/s
-3
0.10      0.10   1.23 x 10
0.10      0.20   2.46 x 10-3
0.20      0.10   4.92 x 10-3

S.D: In this problem we determine the rate expression and rate constant for the reaction, knowing the
rates of the reaction at different concentrations of reactants.

Solution:

a) Rate law for the reaction is,

Rate = k [NO]a [H2]b

Using the given data,

Rate 1 = k [0.10]a [0.10]b = 1.23 x 10-3 -------> (1)
a        b          -3
Rate 2 = k [0.10] [0.20] = 2.46 x 10 -------> (2)

Rate 3 = k [0.20]a [0.10]b = 4.92 x 10-3 -------> (3)

Divide equation (2) by equation (1):

b
2 = (2)

b=1

Divide equation (3) by equation (1):

4 = (2)a

2a = 22
a=2

Therefore the rate expression for the given reaction is,

Rate = k [NO]2 [H2]

b) Recall equation (1).

1.23 x 10-3 M/s = k [0.10]a [0.10]b

Substituting a = 2 and b = 1, we get,

1.23 x 10-3 M/s = k [0.10 M]2[0.10 M]1

Therefore the rate constant for the given reaction is               .
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2. Consider the reaction, H2(g) + I2(g) → 2HI(g). The rate constant for the formation of HI is 2.7 x 10-4
L/(mol.s) at 600 K and 3.5 x 10-3 L/(mol.s) at 650 K. Find the activation energy [Ea].

Solution:

S.D: In this problem we calculate the activation energy knowing the rate constants of the reaction at
two different temperatures.

k1 = 2.7 x 10-4 L/(mol.s)             T1 = 600 K
-3
k2 = 3.5 x 10 L/(mol.s)               T2 = 650 K

Gas constant, R = 8.314 J/(mol.K)
Therefore the activation energy of the reaction is                    .
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3. The half life of a decomposition reaction is 143 s. Determine the value of the rate constant.

S.D: In this problem we determine the rate constant for a reaction from the half life of the reaction.

Solution:

Half life, t1/2 = 143 s

Rate constant ( ) is related to half life (t1/2) as follows:

= 0.00485 s-1
-3 -1
= 4.85 x 10 s
-3 -1
Therefore, rate constant for the reaction is 4.85 x 10 s .
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4. The mechanism of a reaction is shown below:
HOOH + I- → HOI + OH- (slow)

HOI + I- → I2 + OH-           (fast)

2OH- + 2H3O+ → 4H2O (fast)

a) What is the overall reaction?

b) Predict the rate law based on the mechanism.

S.D: In this problem we predict the rate law of a reaction based on the mechanism of the reaction.

Solution:

a) The overall reaction is obtained by adding the three given reactions in the mechanism.
HOOH + I- → HOI + OH-
HOI + I- → I2 + OH-
-           +
2OH + 2H3O → 4H2O
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HOOH + 2I- + 2H3O+ → I2 + 4H2O
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Therefore the overall reaction is,

HOOH + 2I- + 2H3O+ → I2 + 4H2O

b) The rate law for the reaction is determined by the slow step, which in the given case is the first step
of the mechanism.
Therefore, the rate law based on the mechanism is,

Rate = k [HOOH][ I-]
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5. What fraction of A will remain after 150 s if the rate constant of the following first order reaction
A → products, is 2.95 x 10-3 s-1.

S.D: In this problem we determine the fraction of a reactant remaining in a reaction after a certain
time period.

Solution:

Let the initial concentration of A be [A]0

Let the final concentration of A be [A]

Fraction of A remaining after 150 s

For a first order reaction, the integrated rate law is,

Therefore the fraction of A remaining after 150 s is 0.642
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6. The initial concentration of reactants of a second order reaction is 1.00 M and the final concentration
after 25 min is 0.25 M. What is the rate constant for the reaction?

S.D: In this problem we calculate the rate constant of a second order reaction from the initial and final
concentration of reactants.

Solution:

Initial concentration [A]0 = 1.00 M

Final concentration [A] = 0.25 M

Time t = 25 min

The integrated rate law for a second order reaction is,

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7. Express the rate of the following reaction in terms of the concentrations of reactants and products.
2NO2 → 2NO + O2

S.D: In this problem we express the rate of a reaction in terms of concentrations of reactants and
products.

Solution:

The rate for the given reaction may be expressed in terms of rate of disappearance of NO2 or rate of
appearance of NO or O2.
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8. Ammonia and oxygen react at high temperature as:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
In an experiment, rate of formation of NO is 3.6 x 10-3 mol L-1 s-1. Calculate the rate of formation of
water.

S.D: In this problem we calculate the rate of formation of a product from the rate equation.

Solution:

The rate equation for the given reaction is,

From the rate equation,

Therefore, the rate of formation of water
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9. Calculate the order of the reaction which has the following rate equation.

S.D: In this problem we determine order of a reaction from the rate equation.

Solution:
Order of a reaction is defined as the sum of the powers of the concentration terms of the reactants in
the rate equation.

Given:

Therefore, the order of the reaction is 2.
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10. A reaction that is of first order with respect to reactant A has a rate constant 6 min-1. If the initial
concentration of A is 5.0 mol L-1, when would the concentration of A reach the value of 0.05 mol L-1.

S.D: In this problem we determine the time taken by a reactant to reach a particular concentration,
knowing the rate constant of the reaction.

Solution:

Initial concentration [A]0 = 5.0 mol L-1

Final concentration [A] = 0.05 mol L-1

Rate constant k = 6 min-1

Time taken t = ?

For a first order reaction,

Therefore the concentration of A would reach a value of 0.05 mol L-1 in 0.768 min.
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11. A first order reaction is 20% complete in 10 minutes. What is the time taken for 75% completion of
the reaction?
S.D: In this problem we calculate the time taken for a certain percentage of a reaction to be
completed using the first order rate constant expression.

Solution:

First case:
Let the initial concentration of reactant be ‘a’.
Therefore, [A]0 = a
Final concentration, [A] = a - 20 % of a

Time, t = 10 min

For a first order reaction, rate constant is given as,

Second case:

Let the initial concentration of reactant be ‘a’.
Therefore, [A]0 = a
Final concentration, [A] = a - 75 % of a

Rate constant, k = 0.0223 min-1

Time taken, t = ?

For a first order reaction,
Therefore, the time taken for 75% completion of the reaction is 62.18 min.
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12. Identify the reaction order from each of the following rate constants.
(a) k = 2.3 × 10–5 L mol–1 s–1
–4 –1
(b) k = 3 × 10 s

S.D: In this problem we determine the reaction order by knowing the units of rate constant.

Solution:
–1 –1                         –5      –1 –1
(a) The unit of second order rate constant is L mol s , therefore k = 2.3 × 10 L mol s represents a
second order reaction.

(b) The unit of a first order rate constant is s–1 therefore k = 3 × 10–4 s–1 represents a first order reaction.
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13. The conversion of substance X to Y follows second order kinetics. If the concentration of X is tripled,
how will it affect the rate of the reaction?

S.D: In this problem we study the relation between change in concentration of a reactant and the rate
of the reaction.

Solution:

Since the reaction follows second order kinetics, the rate law for the reaction is,
Rate = k [X]2

When the concentration of X is tripled,
2
Rate = k [3X]
= 9{k [X]2}
Therefore, when the concentration of X is tripled, the rate increases 9 times.
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14. The thermal decomposition of a compound is of first order. If the half life of the compound is 120
min, how long will it take for 90% of the compound to decompose?
S.D: In this problem we calculate the time taken for a particular percentage of the reaction to occur by
knowing the half life of the reaction.

Solution:

Half life (t1/2) = 120 min

For a first order reaction, rate constant(k) is given as,

Let the initial concentration of the compound be ‘a’.
Therefore, [A]0 = a
90% of the compound is decomposed.
Therefore final concentration, [A] = a - 90 % of a

Rate constant,

Time taken, t = ?

For a first order reaction,

Therefore, the time taken for 90% of the compound to decompose is 399 min.
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15. Can activation energy for a reaction be zero?
S.D: In this question we prove that the activation energy for a reaction cannot be zero.

Solution:

According to Arrhenius equation,

Where, ‘k’ is the rate constant, ‘A’ is the frequency factor and ‘Ea’ is the activation energy.

If Ea is zero, then according to Arrhenius equation, k= A.
This would mean that every collision between molecules leads to a chemical reaction, which is not true.

Hence, Ea cannot be zero.
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16. Show that for a first order reaction, the time required for 99.9% completion is 10 times that required
for 50% completion.

S.D: In this problem we derive the relation between time taken for 99.9% completion and time taken
for 50% completion of a first order reaction.

Solution:

For a first order reaction,

Time taken for 99.9% completion:
Let the time taken for 99.9% completion of the reaction be

Let initial concentration, [A]0 = a
Then, final concentration, [A] = a – 99.9% of a
Time taken for 50% completion:
Let the time taken for 50% completion of the reaction be

Let initial concentration, [A]0 = a
Then, final concentration, [A] = a – 50% of a

Dividing (1) by (2), we get

Therefore, for a first order reaction, the time required for 99.9% completion is 10 times that required for
50% completion.
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17. The rate constant for a first order reaction becomes six times when the temperature is raised from
350 K to 400 K. Calculate the activation energy for the reaction. (Gas constant R = 8.314 J/(mol.K)).

S.D: In this problem we determine the activation energy of a reaction by knowing the ratio of rate
constants at two different temperatures.

Solution:

T1 = 350 K
Let the rate constant at 350 K be k1
T2 = 400 K
Let the rate constant at 400 K be k2

Given, k2 = 6 k1 (or)

(or)

Therefore, the activation energy for the reaction 41.7 kJ/mol.
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18. Consider the following reaction:
NO2 + CO → NO + CO2
A suggested mechanism for the above reaction is given below:
NO2 + NO2 → NO3 + NO ------------ (step-1)
NO3 + CO → NO2 + CO2 ------------ (step-2)
If the rate law is known to be k[NO2]2,
a) which of the two steps is slower?
b) which substance (if any) is probably an unstable intermediate?

S.D: In this problem we show that only the reactants of the slow step of a reaction are found in the
rate law of the reaction.

Solution:

a) The first step would be slow since all of the ingredients for the rate law are found in the reactants of
the first step.
b) NO3, which is produced in one step, consumed subsequently and does not appear in the net reaction
is an intermediate.
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19. The half life of a second order reaction is 49.5 s. If the initial concentration of the reactant A is equal
to 0.0400 M, what is the second order rate constant?

S.D: In this problem we determine the rate constant of a second order reaction knowing the half life
of the reaction.

Solution:

Initial concentration [A]0 = 0.0400 M

Half life t1/2 = 49.5 s

Second order rate constant k = ?

For a second order reaction,

Therefore, the rate constant of the second order reaction is                            .
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-3    -1
20. The initial rate of a reaction which is second order in A is 2.0 x 10 M s , when the initial
concentration of A is 0.0050 M. What is the second order rate constant?

S.D: In this problem we determine the rate constant of a second order reaction from the initial rate of
the reaction and initial concentration of the reactants.

Solution:
Initial concentration of A, [A] = 0.0050 M

Initial rate = 2.0 x 10-3 M s-1

Second order rate constant k = ?

For a second order reaction,
Therefore, the rate constant of the second order reaction is    .
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