Chapter 16, Problem 16.122P

(a)

Interpretation Introduction

Interpretation:

The rate law in absence of a catalyst has to be determined.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right)\end{array}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

$\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}\text{,where 'm and n' are orders of the reactants}\text{.}$

The term ‘m’ represents the order of reaction with respect to $\left[\text{A}\right]$ and the ‘n’ represent the order of the reaction with respect to $\left[\text{B}\right]$. The order is the exponent in the relationship between rate and reactant concentration and tells about the concentration of reactant that influences rate.

Calculating the concentration of both A and B reactants in order to find the order of the reactants:

In reaction mixture (I):

Concentration of A $\text{=}\left(\text{5}\text{spheres}\right)\left(\frac{\text{0}\text{.10}\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{\text{1}}{\text{0}\text{.50}\text{L}}\right)\text{=}\text{0}\text{.10}\text{mol}\text{/L}\text{A}$

Concentration of B $\text{=}\left(\text{5}\text{spheres}\right)\left(\frac{\text{0}\text{.01}\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{\text{1}}{\text{0}\text{.50}\text{L}}\right)\text{=}\text{0}\text{.10}\text{mol}\text{/L}\text{B}$

Reaction rate of mixture (I) is, $\text{3}\text{.5}\times {10}^{-4}\text{ = k }{\left[\text{0}\text{.10}\right]}^{\text{m}}{\left[\text{0}\text{.10}\right]}^{\text{n}}$

In reaction mixture (II):

Concentration of A $\text{=}\left(8\text{spheres}\right)\left(\frac{\text{0}\text{.01}\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{\text{1}}{\text{0}\text{.50}\text{L}}\right)\text{=}\text{0}\text{.16}\text{mol}\text{/L}\text{A}$

Concentration of B $\text{=}\left(\text{5}\text{spheres}\right)\left(\frac{\text{0}\text{.01}\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{\text{1}}{\text{0}\text{.50}\text{L}}\right)\text{=}\text{0}\text{.10}\text{mol}\text{/L}\text{B}$

Reaction rate of mixture (II) is, $\text{5}\text{.6}\times {10}^{-4}\text{ = k }{\left[\text{0}\text{.16}\right]}^{\text{m}}{\left[\text{0}\text{.10}\right]}^{\text{n}}$

In reaction mixture (III):

Concentration of A $\text{=}\left(8\text{spheres}\right)\left(\frac{\text{0}\text{.01}\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{\text{1}}{\text{0}\text{.50}\text{L}}\right)\text{=}\text{0}\text{.16}\text{mol}\text{/L}\text{A}$

Concentration of B $\text{=}\left(7\text{spheres}\right)\left(\frac{\text{0}\text{.01}\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{\text{1}}{\text{0}\text{.50}\text{L}}\right)\text{=}\text{0}\text{.14}\text{mol}\text{/L}\text{B}$

Reaction rate of mixture (III) is, $\text{5}\text{.6}\times {10}^{-4}\text{ = k }{\left[\text{0}\text{.16}\right]}^{\text{m}}{\left[\text{0}\text{.14}\right]}^{\text{n}}$

Order of reactant is determined as follows,

$\begin{array}{l}\underset{\_}{\text{Find}\text{order}\text{of}\text{the}\text{reactant A:}}\\ \\ \ast \text{ Comparing}\text{first}\text{two}\text{experiments}\text{1and}\text{2,}\\ \\ \text{rate}1\text{ = k }{\left[\text{A}\right]}_1{}^{\text{m}}{\left[\text{B}\right]}_1{}^{\text{n}}\text{, rate1 = 3}\text{.5}\times {\text{10}}^{-4}\text{ mol/L}\\ \text{rate}2\text{=}\text{k }{\left[\text{A}\right]}_2{}^{\text{m}}{\left[\text{B}\right]}_2{}^{\text{n}}\text{, rate2 = 5}{\text{.6×10}}^{\text{-4}}\text{mol/L}\\ \\ \\ \frac{\text{rate2}}{\text{rate}1}\text{=}\frac{\cancel{\text{k}}{\left[\text{A}\right]}_2{}^{\text{m}}{\left[\text{B}\right]}_2{}^{\text{n}}}{\cancel{\text{k}}{\left[\text{A}\right]}_1{}^{\text{m}}{\left[\text{B}\right]}_1{}^{\text{n}}}\\ \\ \frac{\text{5}{\text{.6×10}}^{\text{-4}}\text{mol/L}}{\text{3}\text{.5}\times {\text{10}}^{-4}\text{ mol/L}}\text{=}\frac{{\left(\text{0}\text{.16}\right)}^m\cancel{{\left(\text{0}\text{.10}\right)}^n}}{{\left(\text{0}\text{.10}\right)}^m\cancel{{\left(\text{0}\text{.10}\right)}^n}}\\ \\ \text{1}\text{.6 = }{\left(1.6\right)}^m\Rightarrow {\left(1.6\right)}^1\text{= }{\left(1.6\right)}^m\\ \\ \boxed{\text{m = 1 }}\end{array}$

Thus, the order with respect to reactant A is first order.

$\begin{array}{l}\underset{\_}{\text{Find}\text{order}\text{of}\text{the}\text{reactant B:}}\\ \\ \ast \text{ Comparing}\text{first}\text{two}\text{experiments}2\text{and}3\text{,}\\ \\ \text{rate}3\text{ = k }{\left[\text{A}\right]}_3{}^{\text{m}}{\left[\text{B}\right]}_3{}^{\text{n}}\text{, rate3 = 5}\text{.6}\times {\text{10}}^{-4}\text{ mol/L}\\ \text{rate}2\text{=}\text{k }{\left[\text{A}\right]}_2{}^{\text{m}}{\left[\text{B}\right]}_2{}^{\text{n}}\text{, rate2 = 5}{\text{.6×10}}^{\text{-4}}\text{mol/L}\\ \\ \\ \frac{\text{rate2}}{\text{rate}3}\text{=}\frac{\cancel{\text{k}}{\left[\text{A}\right]}_2{}^{\text{m}}{\left[\text{B}\right]}_2{}^{\text{n}}}{\cancel{\text{k}}{\left[\text{A}\right]}_3{}^{\text{m}}{\left[\text{B}\right]}_3{}^{\text{n}}}\\ \\ \frac{\text{5}{\text{.6×10}}^{\text{-4}}\text{mol/L}}{\text{5}{\text{.6×10}}^{\text{-4}}\text{mol/L}}\text{=}\frac{\cancel{{\left(\text{0}\text{.16}\right)}^m}{\left(\text{0}\text{.14}\right)}^n}{\cancel{{\left(\text{0}\text{.16}\right)}^m}{\left(\text{0}\text{.10}\right)}^n}\\ \\ \text{1}\text{.0 = }{\left(1.4\right)}^n\Rightarrow \log \left(1.0\right)\text{= }\left(n\right)\text{ log}\left(1.4\right)\\ \\ \boxed{\text{n = 0 }}\end{array}$

Thus, the order with respect to reactant B is Zero order.

Therefore, the rate law of reaction in absence of catalyst is, $\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^1{\left[\text{B}\right]}^0\text{ =}k{\left[A\right]}^1.$

(b)

Interpretation Introduction

Interpretation:

The overall reaction order has to be predicted.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right).\end{array}$

$\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}{\left[\text{C}\right]}^{\text{p}}\text{,where 'm, n and p' are orders of the reactants}\text{.}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

$\begin{array}{l}\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}\text{,where m, and n are orders of the reactants}\text{.}\\ \\ \text{Given}\text{reaction rate : =}k{\left[A\right]}^1.\end{array}$

$\begin{array}{l}\underset{\_}{\text{Order}\text{of}\text{the}\text{overall}\text{reaction:}}\\ \\ \text{Total}\text{order}\text{of}\text{reaction = m + n}\\ \\ \text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^1{\left[\text{B}\right]}^0\text{ =}k{\left[A\right]}^1.\end{array}$

The overall order of the given reaction is, FIRST ORDER.

(c)

Interpretation Introduction

Interpretation:

The value of the rate constant has to be calculated.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right).\end{array}$

$\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}{\left[\text{C}\right]}^{\text{p}}\text{,where 'm, n and p' are orders of the reactants}\text{.}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

The rate law of the given reaction is,

$\text{Given}\text{reaction rate : =}k{\left[A\right]}^1.$

Substituting the required values from experiment (I) into the above reaction rate is,

$\begin{array}{l}\text{Experiment 1: }\left[\text{A}\right]\text{ = 0}\text{.100}\text{;}\left[\text{B}\right]\text{ = 0}\text{.100}\text{;}\text{rate = 3}{\text{.5×10}}^{\text{-4}}\text{mol/L}\text{.s}\\ \\ \text{k}\text{=}\frac{{\left[\text{A}\right]}^{\text{1}}}{\left(\text{rate}\right)}\text{= }\frac{\text{3}{\text{.5×10}}^{\text{-4}}\text{mol/L}\text{.s}}{\text{0}\text{.100}\text{mol/L}}\text{=}35\times 10^{-4}{s}^{-1}.\end{array}$

Hence, the value of rate constant is $35\times 10^{-4}{s}^{-1}.$

(d)

Interpretation Introduction

Interpretation:

The gray cubes in the experiment (IV) whether catalytic effect has or not has to be explained.

Concept introduction:

Catalyst: A substance or a compound which promotes the rate of a chemical reaction is said to be catalyst. Catalysts are classified into two types.  They are homogeneous catalysts and heterogeneous catalysts.

Explanation of Solution

The rate law of the given reaction is,

$\text{Given}\text{reaction rate : =}k{\left[A\right]}^1.$

In reaction mixture (IV):

Concentration of A $\text{=}\left(\text{5 spheres}\right)\left(\frac{\text{0}\text{.01}\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{\text{1}}{\text{0}\text{.50}\text{L}}\right)\text{=}\text{0}\text{.10}\text{mol}\text{/L}\text{A}$

Concentration of B $\text{=}\left(8\text{spheres}\right)\left(\frac{\text{0}\text{.01}\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{\text{1}}{\text{0}\text{.50}\text{L}}\right)\text{=}\text{0}\text{.16}\text{mol}\text{/L}\text{B}$

Substituting the required values from experiment (IV) into the above reaction rate is,

$\begin{array}{l}\text{Experiment 4: }\left[\text{A}\right]\text{ = 0}\text{.100}\text{;}\text{rate = 4}{\text{.9×10}}^{\text{-4}}\text{mol/L}\text{.s}\\ \\ \text{k}\text{=}\frac{{\left[\text{A}\right]}^{\text{1}}}{\left(\text{rate}\right)}\text{= }\frac{\text{4}{\text{.9×10}}^{\text{-4}}\text{mol/L}\text{.s}}{\text{0}\text{.100}\text{mol/L}}\text{=}49\times 10^{-4}{s}^{-1}.\end{array}$

Hence, the value of rate constant is $35\times 10^{-4}{s}^{-1}.$ and the reaction rate of experiment (IV) in presence of catalyst is $49\times 10^{-4}{s}^{-1}$.

Therefore, the gray pellets have a catalytic effect. The rate of reaction and the rate constant are greater with the pellets than without.