For a reaction $2 \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) ightarrow \mathrm{N}_{2}…
For a reaction $2 \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) ightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ the rate law is $\mathrm{d} p\left(\mathrm{~N}_{2} \mathrm{O}ight) / \mathrm{d} t=k\left(p_{\mathrm{NO}}ight)^{2}\left(p_{\mathrm{H}_{2}}ight) .$ The halflife of the reaction when $\left(p_{\mathrm{NO}}ight)_{0}=10 \mathrm{mmHg}$ and $\left(p_{\mathrm{H}_{2}}ight)_{0}=1200 \mathrm{mmHg}$ is found to be $830 \mathrm{~s}$. The half-life when $\left(p_{\mathrm{NO}}ight)_{0}=20 \mathrm{mmHg}$ and $\left(p_{\mathrm{H}_{2}}ight)_{0}=1200 \mathrm{mmHg}$ will be
$830 \mathrm{~s}$
$415 \mathrm{~s}$
$1245 \mathrm{~s}$
$208 \mathrm{~s}$
Solution
Since $\left(p_{\mathrm{H}_{2}}ight)>>\left(p_{\mathrm{NO}}ight)_{0}$ the reaction follows second-order kinetics with respect to NO. Thus, if the pressure of NO is doubled, it half-life becomes half.