Question

Calculate the rate constant for decomposition of ammonium nitrate from the following data:
$N{H_4}N{O_2} \to {N_2} + 2{H_2}O$

Time(minutes)	$10$	$25$	$\infty$
Volume of ${N_2}$ in mL	$6.25$	$13.65$	$35.05$

Given that the reaction follows first order kinetics
A. $0.0197\,{\min ^{ - 1}}$
B. $1.97\,{\min ^{ - 1}}$
C. $0.197\,{\min ^{ - 1}}$
D. None of these

Answer 1

The reaction is first order and hence the rate constant will be only dependent on the concentration of the ammonium nitrate. Volume at infinity also signifies the volume of nitrogen gas when the reaction is completed and this can be used in the formula for first order kinetics.

Formula used: $K = \dfrac{1}{t}\ln (\dfrac{{{V_\infty }}}{{{V_\infty } - {V_t}}})$
Where :
$K$ is the rate constant
$t$ is the time taken
${V_\infty }$ is the volume after ∞ time has passed.
${V_t}$ is the volume after time t

Complete step by step answer:
The decomposition reaction for ammonia can be given as follows:
$N{H_4}N{O_2} \to {N_2} + 2{H_2}O$
According to this reaction, on the reactant side, we only have ammonium nitrate. Hence, the rate of the reaction will only depend on the concentration of ammonium nitrate.
It has already been given in the question, which means that the rate of the reaction depends on the concentration on only one species.
Now, the rate constant is as the name suggests, a constant. The value for it does not change if we consider the volume at twenty five minutes or infinity. The rare constant will be the same each time.
The equation to determine the value for rate constant is for a first order reaction is
$K = \dfrac{1}{t}\ln (\dfrac{{{V_\infty }}}{{{V_\infty } - {V_t}}})$
Where :
$K$ is the rate constant
$t$ is the time taken
${V_\infty }$ is the volume after ∞ time has passed.
${V_t}$ is the volume after time t

So, let us take the values as given in the question.
We will use the value for $t$ as ten minutes since it is easier to use in calculation.
Now,
${V_\infty } = 35.05\,,\,{V_t} = 6.25\,,\,t = 10$
Substituting these values in the above equation we get:
$K = \dfrac{1}{{10}}\ln (\dfrac{{35.05}}{{35.05 - 6.25}})$

Solving the above equation, we get:
$K = 0.0197\,{\min ^{ - 1}}$

So, the correct answer is Option A.

Note: Since the reaction is first order, it didn’t necessarily matter if the volume was taken at ten minutes or twenty five minutes, since both these volumes were a function of the concentration of ammonium nitrate only.
${V_\infty }$ in practicality is determined by leaving the solution mixture overnight and doing titration the next morning. It is presumed that the reaction is completed overnight. This is also a fixed value for a given concentration of reactant.