# The following reaction rate data was obtained in four separate experiments. What is the rate law...

## Question:

The following reaction rate data were obtained in four separate experiments.

 Experiment ( NO₂)₀ M (CO)₀ M Initial rate , Ms⁻¹ 1 1.50 0.490 1.36 2 3.00 0.490 5.45 3 1.50 0.980 1.36 4 3.00 0.980 5.45

What is the rate law for the reaction and what is the numerical value of k?

## Rate Law:

The rate law for a reaction is an expression that expresses the dependence of the reaction rate on the concentrations of different species present in the reaction.

The sum of powers raised to concentration terms in the rate law expression is known as the overall order of the reaction.

The rate law expression for the given reaction may be written as:

{eq}Rate=k(NO_{2})^{x}(CO)^{y} {/eq}

where

'x' and 'y' are orders w.r.t. {eq}NO_{2} {/eq} and CO, respectively.

From the given data:

{eq}1.36M/s=k(1.50M)^{x}(0.490M)^{y}-----(1) {/eq}

{eq}5.45M/s=k(3.00M)^{x}(0.490M)^{y}-----(2) {/eq}

{eq}1.36M/s=k(1.50M)^{x}(0.980M)^{y}-----(3) {/eq}

Dividing equation (1) by (3):

{eq}\dfrac{1.36M/s=k(1.50M)^{x}(0.490M)^{y}}{1.36M/s=k(1.50M)^{x}(0.980M)^{y}} {/eq}

{eq}\Rightarrow y=0 {/eq}

Dividing equation (1) by (2):

{eq}\dfrac{1.36M/s=k(1.50M)^{x}(0.490M)^{y}}{5.45M/s=k(3.00M)^{x}(0.490M)^{y}} {/eq}

{eq}\Rightarrow x=2 {/eq}

Thus, the rate of the reaction is:

{eq}Rate=k(NO_{2})^{2}(CO)^{0} {/eq}

Using equation (1):

{eq}1.36M/s=k(1.50M)^{2}(0.490M)^{0} {/eq}

{eq}\Rightarrow k=0.604M^{-1}s^{-1} {/eq} 