Question

$N_2{O}_5\left(g\right)\to 2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

In the above first order reaction the initial concentration of $N_2{O}_5$ is 2.40 $\times {10}^{-2}mol{L}^{-1}$ at 318 K. The concentration of $N_2{O}_5$ after 1 hour was $1.60\times {10}^{-2}mol{L}^{-1}$ . The rate constant of the reaction at 318 k is $x\times {10}^{-3}mi{n}^{-1}$ The value of x (Nearest integer) is

$[Given:log3=0.447,log5=0.699]$




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Answer 1

For the first order reaction

$kt=\ln \frac{{\left[R\right]}_0}{\left[R\right]}$
$\left[R_0\right]$ is the initial concentration
$\left[R\right]$ is the final concentration at time t

$k\times 60=\ln \frac{(2.4\times {10}^{-2})}{(1.6\times {10}^{-2})}$

$k=2.303\times (log3-log2)$

$k=\frac{2.303}{60}\times (0.477-0.301)$

$k=6.7\times {10}^{-3}mi{n}^{-1}$
$k\approx 7\times {10}^{-3}mi{n}^{-1}x=7$