Question

The following data were collected for the rate of disappearance ofNO in the reaction 2NO(g) + O₂ --> 2NO₂(g) :

	Experiment	[NO] (M)	[O2] (M)	Initial Rate (M/s)
	1	0.0126	0.0125	1.41 x 10-2
	2	0.0252	0.0125	5.64 x 10-2
	3	0.0252	0.0250	1.13 x 10 -1

(a) What is the rate law for the reaciton? (B) What are teh unitsof the rate constant? (c) What is the average value of the rateconstant calculated form teh three data sets? (d) What is the rateof disappearance of NO when [NO] = 0.0750 M and [O₂] =0.0100 M? (e) What is the rate of disappearance of O₂ atteh concentarations given in part (d)?

Answer 1

Concepts and reason

The rate of the reaction for a particular reaction explains how slowly or quickly a reaction takes place. The rate law can be used to find the rate of a reaction.

The order of a reaction is defined as an index to which a concentration term is raised in the rate law. The order of a reaction is found using experimental results. By substituting the experimental results for a reaction and the concentration of reactants the order of the reaction can be calculated.

Fundamental

For the reaction,

${\rm{A}} + {\rm{B}} \to {\rm{C}} + {\rm{D}}$

The rate of the reaction is written as follows:

$r = k{\left[ A \right]^x}{\left[ B \right]^y}$

Here, $\left[ A \right]$ , $\left[ B \right]$ are the molar concentration of reactants A and B respectively. And x and y are order of the reaction with respect to A and B.

(1)

The reaction is as follows:

${\rm{2NO}} + {{\rm{O}}_2} \to 2{\rm{N}}{{\rm{O}}_{\rm{2}}}$

The rate law is written as follows:

$r = k{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]^x}{\left[ {{{\rm{O}}_2}} \right]^y}$

From experiment 1 and 2

$1.41 \times {10^{ - 2}} = k{\left[ {0.0126} \right]^x}{\left[ {0.0125} \right]^y}$ …… (1)

And

$5.64 \times {10^{ - 2}} = k{\left[ {0.0252} \right]^x}{\left[ {0.0125} \right]^y}$ …… (2)

Divide equation (2) by (1) as follows:

$4 = {\left[ 2 \right]^x}$

On solving,

$x = 2$

From experiment 2 and 3

$5.64 \times {10^{ - 2}} = k{\left[ {0.0252} \right]^x}{\left[ {0.0125} \right]^y}$ …… (2)

And

$1.13 \times {10^{ - 1}} = k{\left[ {0.0252} \right]^x}{\left[ {0.0250} \right]^y}$ …… (3)

Divide equation (3) by (2) as follows:

$2.00 = {\left[ 2 \right]^{\rm{y}}}$

On solving,

$y = 1$

Thus, the rate law is written as follows:

$r = k{\left[ {{\rm{NO}}} \right]^2}{\left[ {{{\rm{O}}_{\rm{2}}}} \right]^1}$

(2)

The rate law of the reaction is as follows:

$r = k{\left[ {{\rm{NO}}} \right]^2}{\left[ {{{\rm{O}}_{\rm{2}}}} \right]^1}$

Substituting ${\rm{M}}/{\rm{s}}$ for the units of r and M for $\left[ {{\rm{NO}}} \right]$ and $\left[ {{{\rm{O}}_{\rm{2}}}} \right]$ .

$\begin{array}{c}\\{\rm{M}}/{\rm{s}} = k{\left[ {\rm{M}} \right]^2}{\left[ {\rm{M}} \right]^1}\\\\k = \frac{{{\rm{M}}/{\rm{s}}}}{{{{\rm{M}}^3}}}\\\\k = {{\rm{M}}^{ - 2}}{{\rm{s}}^{ - 1}}\\\end{array}$

(3)

The rate law of the given reaction is as follows:

$\begin{array}{c}\\r = k{\left[ {{\rm{NO}}} \right]^2}{\left[ {{{\rm{O}}_{\rm{2}}}} \right]^1}\\\\k = \frac{r}{{{{\left[ {{\rm{NO}}} \right]}^2}{{\left[ {{{\rm{O}}_{\rm{2}}}} \right]}^1}}}\\\end{array}$ ……(4)

For experiment (1), the rate is written as follows:

$\begin{array}{c}\\{k_1} = \frac{{1.41 \times {{10}^{ - 2}}}}{{{{\left[ {0.0126} \right]}^2}{{\left[ {0.0125} \right]}^1}}}\\\\ = 7105.064{\rm{ }}{{\rm{M}}^{ - 2}}{{\rm{s}}^{ - 1}}\\\end{array}$

For experiment 2, the rate law is as follows:

$\begin{array}{c}\\{k_2} = \frac{{5.64 \times {{10}^{ - 2}}}}{{{{\left[ {0.0252} \right]}^2}{{\left[ {0.0125} \right]}^1}}}\\\\ = 7105.064{\rm{ }}{{\rm{M}}^{ - 2}}{{\rm{s}}^{ - 1}}\\\end{array}$

For experiment 3, the rate law is as follows

$\begin{array}{c}\\{k_3} = \frac{{1.13 \times {{10}^{ - 1}}}}{{{{\left[ {0.0252} \right]}^2}{{\left[ {0.0250} \right]}^1}}}\\\\ = 7117.662{\rm{ }}{{\rm{M}}^{ - 2}}{{\rm{s}}^{ - 1}}\\\end{array}$

(3)

Take the average of all the rate constant as follows:

$\begin{array}{c}\\{k_{avg}} = \frac{{7105.064 + 7105.064 + 7117.662}}{3}\\\\ = 7109.293{\rm{ }}{{\rm{M}}^{ - 2}}{{\rm{s}}^{ - 1}}\\\end{array}$

(4)

The rate law of the given reaction is:

$r = k{\left[ {{\rm{NO}}} \right]^2}{\left[ {{{\rm{O}}_{\rm{2}}}} \right]^1}$

Substitute $0.0750{\rm{ M}}$ for $\left[ {{\rm{NO}}} \right]$ , $0.0100{\rm{ M}}$ for $\left[ {{{\rm{O}}_{\rm{2}}}} \right]$ and $7109.293{\rm{ }}{{\rm{M}}^{ - 2}}{{\rm{s}}^{ - 1}}$ for k.

$\begin{array}{c}\\r = \left( {7109.293{\rm{ }}{{\rm{M}}^{ - 2}}{{\rm{s}}^{ - 1}}} \right){\left[ {0.0750{\rm{ M}}} \right]^2}\left[ {0.0100{\rm{ M}}} \right]\\\\ = 0.3998\\\\ \approx 0.4{\rm{ M}}/s\\\end{array}$

(5)

The rate of disappearance of ${{\rm{O}}_2}$ is:

$- \frac{{d\left[ {{{\rm{O}}_2}} \right]}}{{dt}} = - \frac{1}{2}\frac{{d\left[ {{\rm{NO}}} \right]}}{{dt}}$

Substitute $0.4{\rm{ M}}/s$ for $\frac{{d\left[ {{\rm{NO}}} \right]}}{{dt}}$ .

$\begin{array}{c}\\\frac{{d\left[ {{{\rm{O}}_2}} \right]}}{{dt}} = \frac{1}{2} \times 0.4\\\\ = 0.2{\rm{ M}}/{\rm{s}}\\\end{array}$

Ans: Part 1

The rate law of the reaction is as follows:

$r = k{\left[ {{\rm{NO}}} \right]^2}{\left[ {{{\rm{O}}_{\rm{2}}}} \right]^1}$

Part 2

The units of rate constant is ${{\rm{M}}^{ - 2}}{{\rm{s}}^{ - 1}}$ .

Part 3

The average value of rate constant is $7109.293{\rm{ }}{{\rm{M}}^{ - 2}}{\rm{ }}{{\rm{s}}^{ - 1}}$ .

Part 4

The rate of disappearance of NO is $0.4{\rm{ M}}/s$ .

Part 5

The rate of disappearance of ${{\rm{O}}_2}$ is $0.2{\rm{ M}}/{\rm{s}}$ .