The gaseous reaction A(g)→2B(g)+C(g) is found to be first order. If the reaction is started with pA= 90 mmHg , the total pressure after 10 min is found to be 180 mmHg. The rate constant of the reaction is

# The gaseous reaction $\mathrm{A}\left(\mathrm{g}\right)\to 2\mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{g}\right)$ is found to be first order. If the reaction is started with  , the total pressure after 10 min is found to be . The rate constant of the reaction is

1. A

$1.15×{10}^{-3}{\mathrm{s}}^{-1}$

2. B

$2.30×{10}^{-3}{\mathrm{s}}^{-1}$

3. C

$3.45×{10}^{-3}{\mathrm{s}}^{-1}$

4. D

$4.60×{10}^{-3}{\mathrm{s}}^{-1}$

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### Solution:

$\underset{{\mathrm{p}}_{0}-\mathrm{p}}{\mathrm{A}\left(\mathrm{g}\right)}\to \underset{2\mathrm{p}}{2\mathrm{B}\left(\mathrm{g}}\right)+\underset{\mathrm{p}}{\mathrm{C}\left(\mathrm{g}\right)}$

Total pressure $=\left({p}_{0}-p\right)+\left(2p\right)+p={p}_{0}+2p$

or

Now, $\mathrm{log}\frac{{p}_{0}-p}{{p}_{0}}=-\frac{k}{2.303}t$

$\mathrm{log}\frac{45}{90}=-\frac{k}{2,303}\left(10×60\mathrm{s}\right)$. Thus   $k=\frac{0.301×2.303}{10×60\mathrm{s}}=1.155×{10}^{-3}{\mathrm{s}}^{-1}$

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