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Chemical Kinetics and Nuclear Chemistry question

2022 · 25 Jul · Shift 1 · Q15

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Chemical Kinetics and Nuclear Chemistry
2022 · 25 Jul · Shift 1 · Q15

Chemical Kinetics and Nuclear Chemistry question

2022 · 25 Jul · Shift 1 · Q15

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1

The half life for the decomposition of gaseous compound

\mathrm{A}

is

240 \mathrm{~s}

when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be

4.0

min. The order of the reaction is

\underline{\hspace{2cm}}

. (Nearest integer)

Numerical answer

View written solutionFree

Correct answer: 1

Use half-life dependence on initial concentration/pressure

For a gas at constant temperature, concentration is proportional to pressure. So we can use initial pressure in place of initial concentration.

For an $n$ -th order reaction ( $n \neq 1$ ), the half-life is:

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

where $a$ is the initial concentration (or initial pressure here).

So,

t_{1/2} \propto a^{-(n-1)}

Write the given data

When initial pressure $P_1 = 500$ Torr, half-life $t_1 = 240$ s
When initial pressure $P_2 = 250$ Torr, half-life $t_2 = 4.0$ min $= 240$ s

Thus,

t_1 = t_2

Apply proportionality

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

Since $t_1=t_2$ ,

1 = \left(\frac{250}{500}\right)^{n-1}

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

This is possible only if

n-1 = 0

So,

n = 1

Conclusion

The reaction is first order.

$\boxed{1}$

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