JEE MAIN - Chemistry (2022 - 25th June Evening Shift - No. 18)
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 ____________. [integer answer]
Answer
0
Explanation
$t_{1 / 2} \propto \frac{1}{\left[P_{0}\right]^{n-1}}$
$$ \begin{aligned} &\frac{\left(\mathrm{t}_{1 / 2}\right)_{1}}{\left(\mathrm{t}_{1 / 2}\right)_{2}}=\frac{\left[\mathrm{P}_{0}\right]_{2}^{\mathrm{n}-1}}{\left[\mathrm{P}_{0}\right]_{1}^{\mathrm{n}-1}} \\\\ &\frac{340}{170}=\left(\frac{27.8}{55.5}\right)^{\mathrm{n}-1} \\\\ &2=\left(\frac{1}{2}\right)^{\mathrm{n}-1} \\\\ &2=(2)^{1-n} \\\\ &1-\mathrm{n}=1 \\\\ &\mathrm{n}=0 \end{aligned} $$
$$ \begin{aligned} &\frac{\left(\mathrm{t}_{1 / 2}\right)_{1}}{\left(\mathrm{t}_{1 / 2}\right)_{2}}=\frac{\left[\mathrm{P}_{0}\right]_{2}^{\mathrm{n}-1}}{\left[\mathrm{P}_{0}\right]_{1}^{\mathrm{n}-1}} \\\\ &\frac{340}{170}=\left(\frac{27.8}{55.5}\right)^{\mathrm{n}-1} \\\\ &2=\left(\frac{1}{2}\right)^{\mathrm{n}-1} \\\\ &2=(2)^{1-n} \\\\ &1-\mathrm{n}=1 \\\\ &\mathrm{n}=0 \end{aligned} $$
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