At $298 \mathrm{K},1 \mathrm{atm}$, among
A) $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$
B) ${\mathrm{H}}_2+{\mathrm{Cl}}_2\to 2\mathrm{HCl}$
C) ${\mathrm{N}}_2+{\mathrm{O}}_2\to 2\mathrm{NO}$
D) ${\mathrm{H}}_2{\mathrm{SO}}_4+\mathrm{KOH}\to$ products
Correct order of reaction rates is:

Reaction D is a neutralization reaction between $H_{2}S{O}_4$ and $KOH$ forming $K_{2}S{O}_4$ and $H_{2}O$. It is a very fast reaction and its rate increases.

Reaction B is a fast reaction as $H_2$ and $C{l}_2$ forms $HCl$.

Reaction A is a moderate reaction because $H_2$ and $O_2$ combining to form $H_{2}O$.

Reaction C is a slow reaction because reaction between $N_2$ and $O_2$ forming $NO$ is an endothermic reaction.

So, the order will be:

D > B > A > C

Therefore, the correct option is (C).