Question

2 NO + O22 NO2 is second order in NO and first order in O2. Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n... , where '1' is understood for m, n ... (don't enter 1) and concentrations taken to the zero power do not appear. Rate = In an experiment to determine the rate law, the rate constant was determined to be 9.87×103 M-2s-1. Using this value for the rate constant, the rate of the reaction when [NO] = 7.86×10-3 M and [O2] = 2.21×10-3 M would be Ms-1.

Answer 1

Answer : The value of rate of reaction is $1.35\times 10^{-8}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given chemical equation is:

$2NO+O_2\rightarrow 2NO_2$

Rate law expression for the reaction is:

$\text{Rate}=k[NO]^a[O_2]^b$

As per question,

a = order with respect to $NO$  = 2

b = order with respect to $O_2$ = 1

Thus, the rate law becomes:

$\text{Rate}=k[NO]^2[O_2]^1$

Now, calculating the value of rate of reaction by using the rate law expression.

Given :

k = rate constant = $9.87\times 10^3M^{-2}s^{-1}$

[NO] = concentration of NO = $7.86\times 10^{-3}M$

$[O_2]$ = concentration of $O_2$= $2.21\times 10^{-3}M$

Now put all the given values in the above expression, we get:

$\text{Rate}=(9.87\times 10^3M^{-2}s^{-1})\times (7.86\times 10^{-3}M)^2\times (2.21\times 10^{-3}M)^1$

$\text{Rate}=1.35\times 10^{-8}Ms^{-1}$

Hence, the value of rate of reaction is $1.35\times 10^{-8}Ms^{-1}$