Question

The oxidation of NO by O3 is first order in each of the reactants, and its rate constant is 1.5 x 107 M.S1. If the concentrations of NO and Os are each 5.0 x 107 M, what is the rate of oxidation of NO in M.s1? a. 3.8 x 10 b. 2.5x 1014 c. 7.5 x 10 d. 15 e. 7.5

Answer 1

Answer:

Rate of oxidation of NO = $3.8 \times 10^{-6} Ms^{-1}$

Explanation:

$O_3$ reacts with NO to form $NO_2$ and $O_2$

$NO + O_3 \rightarrow NO_2 + O_2$

For this reaction,

Rate of reaction = $-\frac{dNO}{dt} =- \frac{dO_3}{dt}$

Minus sign is written because NO and $O_3$ are disappearing in the reaction.

So rate of reaction is equal to rate of oxidation of NO.

Reaction is first order with respect to each reactant, So,

$Rate = k [NO][O_3]$

Where, k = Rate constant

k = $1.5 \times 10^7\ Ms^{-1}$

$[NO] = 5.0 \times 10^7\ M$

$[O_3]=5.0 \times 10^7\ M$

$Rate = k [NO][O_3]$

$=1.5 \times 10^7 \times (5.0 \times 10^7)^2 = 3.75 \times 10^{-6} Ms^{-1} \approx 3.8 \times 10^{-6} Ms^{-1}$

So, rate of oxidation of NO = $3.8 \times 10^{-6} Ms^{-1}$