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Chemical Kinetics and Nuclear Chemistry question

2022 · 25 Jun · Shift 2 · Q18

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Chemical Kinetics and Nuclear Chemistry
2022 · 25 Jun · Shift 2 · Q18

Chemical Kinetics and Nuclear Chemistry question

2022 · 25 Jun · Shift 2 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1

At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.8 kPa, the half life was found to be 170 s. The order of the reaction is

\underline{\hspace{2cm}}

. [integer answer]

Numerical answer

View written solutionFree

Correct answer: 0

Use the relation between half-life and initial concentration/pressure

For a reaction of order $n$ (with $n \neq 1$ ), the half-life is given by

$t_{1/2} \propto \frac{1}{a^{\,n-1}}$

where $a$ is the initial concentration. For a gas at constant temperature,

$a \propto P_0$

So,

$t_{1/2} \propto \frac{1}{P_0^{\,n-1}}$

Write ratio of the two half-lives

Given:

At $P_1 = 55.5\,\text{kPa}$ , $t_1 = 340\,\text{s}$
At $P_2 = 27.8\,\text{kPa}$ , $t_2 = 170\,\text{s}$

Thus,

$\frac{t_1}{t_2} = \left(\frac{P_2}{P_1}\right)^{n-1}$

Substitute values:

$\frac{340}{170} = \left(\frac{27.8}{55.5}\right)^{n-1}$

$2 = \left(\frac{1}{2}\right)^{n-1}$

Now,

$2 = 2^{-\,(n-1)}$

So,

$1 = -(n-1)$

$n-1 = -1$

$n = 0$

Conclusion

The reaction is zero order.

$\boxed{0}$

Comparison with stored answer

Stored correct answer = $0$

This matches the derived answer.

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