Chapter 11, Problem 11.33PAE

The following experimental data were obtained for the reaction of \'I1₄* and NOf in acidic solution.

NH/(aq) + NO₂-(aq) — N_;(g) + 2 H,O(f)

INH/I (mol L1)	[NO21 (mol L-1,	Rate = A[NJ/At (mol L-1 s’)
0.0092	0.098	3.33 X IO"7
0.0092	0.049	1.66 X 10‘7
0.0488	0.196	3.51 X 10"6
0.0249	0.196	1.80 X 10-6

Determine the rate law for this reaction and calculate the rate constant.

Interpretation Introduction

Interpretation: Given the experimental data obtained for a reaction, determine the rate law for the reaction and the value of rate constant

Concept Introduction: Orders of reaction are constantly determined by doing experiments. Consequently without experimental information, we can't conclude anything about the order of a reaction just by having a look at the equation for the reaction. By doing experiments involving a reaction between A and B, the rate of the reaction is identified to be related to the concentrations of A and B as follows:

$rate=k{[A]}^a{[B]}^b$ ->

This is the Rate Equation.

Where,

Rate is in the units of mol dm-³s-1

k is the rate constant

A, B- concentrations in mol dm-3

a - Order of reaction with respect to A

b- Order of reaction with respect to B

Answer to Problem 11.33PAE

Solution: The rate law of the reaction is $Rate=k{[N{H}_4{}^{+}]}^1{[N{O}_2{}^{-}]}^1$ and the rate constant k is $6.93\times {10}^{-3}{M}^{-1}{s}^{-1}$.

Explanation of Solution

Given information: Reaction of $N{H}_4{}^{+}$ and $N{O}_2{}^{-}$ in acidic solution: $N{H}_4{}^{+}(aq)+N{O}_2{}^{-}(aq)\to N_2(g)+2H_{2}O(l)$

Experimental Data

Step 1: For the reaction:

$N{H}_4{}^{+}(aq)+N{O}_2{}^{-}(aq)\to N_2(g)+2H_{2}O(l)$

The rate law can be determined using the rate equation as follows:

$Rate=k{\left[N{H}_4{}^{+}\right]}^a{\left[N{O}_2{}^{-}\right]}^b$

Where,

a= Order of the reaction with respect to $N{H}_4{}^{+}$

b= Order of the reaction with respect to $N{O}_2{}^{-}$

Step 2: From the first and second rows of the given experimental data,

$N{H}_4{}^{+}$ =constant, So. $rate=k{[N{O}_2{}^{-}]}^b$

$\begin{array}{l}k{\left(0.098\right)}^b =\text{ 3}\text{.33}\times {\text{10}}^{-7}\to (1)\hfill \\ k{\left(0.049\right)}^b =\text{ 1}\text{.66}\times {\text{10}}^{-7}\to (2)\hfill \end{array}$

Step 3: Divide (1) by (2), we get

$\begin{array}{l}\begin{array}{l}\frac{3.33\times {10}^{-7}}{1.66\times {10}^{-7}}=\frac{k{(0.098)}^b}{k{(0.049)}^b}\\ 2 ={\left(2\right)}^b\end{array}\hfill \\ b=1\hfill \end{array}$

Step 4: From the third and fourth rows of the given experimental data,

$N{O}_2{}^{-}$ =constant, So $rate=k{[N{H}_4{}^{+}]}^a$

$\begin{array}{l}k{\left(0.0488\right)}^a =\text{ 3}\text{.51}\times {\text{10}}^{-6}\to (3)\hfill \\ k{\left(0.0249\right)}^a =\text{ 1}\text{.80}\times {\text{10}}^{-6}\to (4)\hfill \end{array}$

Step 5: Divide (3) by (4), we get

$\begin{array}{l}\begin{array}{l}\frac{3.51\times {10}^{-6}}{1.80\times {10}^{-6}}=\frac{k{(0.0488)}^a}{k{(0.0249)}^a}\\ 1.95 ={\left(1.95\right)}^a\end{array}\hfill \\ a=1\hfill \end{array}$

Step 6: Rate Equation = >$Rate=k{\left[N{H}_4{}^{+}\right]}^a{\left[N{O}_2{}^{-}\right]}^b =Rate=k{\left[N{H}_4{}^{+}\right]}^1{\left[N{O}_2{}^{-}\right]}^1$

To determine k, pick equation (1)

$\begin{array}{l}3.33\times {10}^{-7}M{s}^{-1} =k\left(0.0092\right)\left(0.098\right)\hfill \\ k=\text{ 6}\text{.93}\times {\text{10}}^{-3}{M}^{-1}{s}^{-1}\hfill \end{array}$

Conclusion

It does not make a difference what the number of reactants there are. The concentration of every reactant will be present in the rate equation, raised to some power. These powers resemble the individual orders with respect to each reactant. The sum of these powers results in the overall order of the reaction. The rate constant will be a constant value for a given reaction only if the concentration of the reactants is changed without changing any other factors.