Given that the mechanism to the left and the steady$-$state approximation results in the following rate law:

Rate $=\frac{k_1{k}_2[NO{]}^2[O_2]}{k_1+k_2[O_2]}$

what would the rate law be if the concentration of $O_2$ was very very very small?

Rate $=k_1[NO{]}^2$

Rate $=k_1{k}_2[NO][O_2]$

Rate $=k_1{k}_2[NO{]}^2$

Rate $=\frac{k_1{k}_2[NO{]}^2}{k_1+k_2[O_2]}$

Rate $=\frac{k_1{k}_2[O_2]}{k_1+k_2[O_2]}$

Rate $=\frac{k_1{k}_2}{k_{-1}}[NO{]}^2[O_2]$

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