Chapter 16, Problem 16.88P

(a)

Interpretation Introduction

Interpretation:

The reaction order with respect to each reactant has to be found.

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

The reaction rate of the chemical reaction:

$\begin{array}{l}\text{Given}\text{reaction: 2A + 2B}+\text{C}\to \text{D + 3E}\\ \\ \text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}{\left[\text{C}\right]}^p\text{,where 'm, n and p' are orders of the reactants}\text{.}\end{array}$

Let’s find the order of reactant $\left[\text{A}\right]$ by comparing first and second experiments as follows,

$\begin{array}{l}\underset{\_}{\text{Find}\text{order}\text{of}\text{the}\text{reaction:}}\\ \\ \ast \text{ Comparing}\text{first}\text{two}\text{experiments}\text{1}\text{and}\text{2,}\\ \\ \text{rate}\text{1}\text{=}\text{k }{\left[\text{A}\right]}_{\text{1}}^{\text{m}}{\left[\text{B}\right]}_{\text{1}}^{\text{n}}{\left[\text{C}\right]}_{\text{1}}^p\text{, rate1 = 6}{\text{.0 ×10}}^{\text{-6}}\text{mol/L}\text{.s}\\ \text{rate}\text{2 = k}{\left[\text{A}\right]}_2^{\text{m}}{\left[\text{B}\right]}_2^{\text{n}}{\left[\text{C}\right]}_2^p\text{, rate2 = 9}\text{.6}\times {\text{10}}^{-5}\text{ mol/L}\text{.s}\\ \\ \frac{\text{rate}2}{\text{rate}1}\text{=}\frac{\cancel{\text{k}}{\left[\text{A}\right]}_2^{\text{m}}{\left[\text{B}\right]}_2^{\text{n}}{\left[\text{C}\right]}_2^{\text{p}}}{\cancel{\text{k}}{\left[\text{A}\right]}_{\text{1}}^{\text{m}}{\left[\text{B}\right]}_{\text{1}}^{\text{n}}{\left[\text{C}\right]}_{{}^{\text{1}}}^{\text{p}}}\\ \\ \frac{\text{9}\text{.6}\times {\text{10}}^{-5}\text{ mol/L}\text{.s}}{\text{6}{\text{.0 ×10}}^{\text{-6}}\text{mol/L}\text{.s}}\text{=}\frac{{\left(0.096\right)}^m\bcancel{{\left(0.085\right)}^n}\bcancel{{\left(0.032\right)}^p}}{{\left(0.024\right)}^m\bcancel{{\left(0.085\right)}^n}\bcancel{{\left(0.032\right)}^p}}\\ \\ \text{16 = }{\left(4\right)}^m\Rightarrow {\left(4\right)}^2\text{= }{\left(4\right)}^m\\ \\ \boxed{\text{m = 2 }}\end{array}$

Let’s find the order of reactant $\left[\text{B}\right]$ by comparing second and fourth experiments as follows,

$\begin{array}{l}\underset{\_}{\text{Find}\text{order}\text{of}\text{the}\text{reaction:}}\\ \\ \ast \text{ Comparing}\text{first}\text{two}\text{experiments}2\text{and}4\text{,}\\ \\ \text{rate}2\text{=}\text{k }{\left[\text{A}\right]}_2^{\text{m}}{\left[\text{B}\right]}_2^{\text{n}}{\left[\text{C}\right]}_2^{\text{p}}\text{, rate2 = 9}{\text{.6 ×10}}^{\text{-5}}\text{mol/L}\text{.s}\\ \text{rate}4\text{ = k}{\left[\text{A}\right]}_4^{\text{m}}{\left[\text{B}\right]}_4^{\text{n}}{\left[\text{C}\right]}_4^{\text{p}}\text{, rate4 =1}\text{.5}\times {\text{10}}^{-6}\text{ mol/L}\text{.s}\\ \\ \frac{\text{rate}2}{\text{rate}4}\text{=}\frac{\cancel{\text{k}}{\left[\text{A}\right]}_2^{\text{m}}{\left[\text{B}\right]}_2^{\text{n}}{\left[\text{C}\right]}_2^{\text{p}}}{\cancel{\text{k}}{\left[\text{A}\right]}_4^{\text{m}}{\left[\text{B}\right]}_4^{\text{n}}{\left[\text{C}\right]}_4^{\text{p}}}\\ \\ \text{substitute}\text{the}\text{value}\text{of}\text{m}\text{=2,}\\ \\ \frac{\text{9}{\text{.6 ×10}}^{\text{-5}}\text{mol/L}\text{.s}}{\text{1}\text{.5}\times {\text{10}}^{-6}\text{ mol/L}\text{.s}}\text{=}\frac{{\left(0.096\right)}^2{\left(0.085\right)}^n\bcancel{{\left(0.032\right)}^p}}{{\left(0.012\right)}^2{\left(0.170\right)}^n\bcancel{{\left(0.032\right)}^p}}\\ \\ \text{64 = }{\left(8\right)}^2{\left(0.5\right)}^n\Rightarrow {\left(1\right)}^2\text{= }{\left(0.5\right)}^n\text{by}\text{taking}\text{lan,}\\ \\ lan\left(1\right)=lan\left[{\left(0.5\right)}^n\right]\Rightarrow 0=n\ln \left(0.5\right)\Rightarrow n=0\\ \\ \boxed{\text{n = 0 }}\end{array}$

Let’s find the order of reactant $\left[\text{C}\right]$ by comparing third and fourth experiments as follows,

$\begin{array}{l}\underset{\_}{\text{Find}\text{order}\text{of}\text{the}\text{reaction:}}\\ \\ \ast \text{ Comparing}\text{first}\text{two}\text{experiments}1\text{and}3\text{,}\\ \\ \text{rate}3\text{=}\text{k }{\left[\text{A}\right]}_3^{\text{m}}{\left[\text{B}\right]}_3^{\text{n}}{\left[\text{C}\right]}_3^p\text{, rate3 = 1}{\text{.5 ×10}}^{\text{-5}}\text{ mol/L}\text{.s}\\ \text{rate}1\text{ = k}{\left[\text{A}\right]}_1^{\text{m}}{\left[\text{B}\right]}_1^{\text{n}}{\left[\text{C}\right]}_1^p\text{, rate1 = 6}\text{.0}\times {\text{10}}^{-6}\text{ mol/L}\text{.s}\\ \\ \frac{\text{rate}1}{\text{rate}3}\text{=}\frac{\cancel{\text{k}}{\left[\text{A}\right]}_1^{\text{m}}{\left[\text{B}\right]}_1^{\text{n}}{\left[\text{C}\right]}_1^p}{\cancel{\text{k}}{\left[\text{A}\right]}_3^{\text{m}}{\left[\text{B}\right]}_3^{\text{n}}{\left[\text{C}\right]}_3^p}\\ \\ \frac{\text{6}\text{.0}\times {\text{10}}^{-6}\text{ mol/L}\text{.s}}{\text{1}\text{.5}\times {\text{10}}^{-5}\text{ mol/L}\text{.s}}\text{=}\frac{\cancel{{\left(0.024\right)}^2}{\left(0.085\right)}^0{\left(0.032\right)}^p}{\cancel{{\left(0.024\right)}^2}{\left(0.034\right)}^0{\left(0.080\right)}^p}\\ \\ \text{Anything}\text{power}\text{zero}\text{is}\text{'1'}\text{.}\text{hence}\text{the}\text{term}\text{'B'}\text{is}\text{ignored}\text{.}\\ \\ \frac{\text{6}\text{.0}\times {\text{10}}^{-6}\text{ mol/L}\text{.s}}{\text{1}\text{.5}\times {\text{10}}^{-5}\text{ mol/L}\text{.s}}\text{=}\frac{{\left(0.032\right)}^p}{{\left(0.080\right)}^p}\\ \\ \text{0}\text{.4 =}{\left(0.4\right)}^p\\ \\ \boxed{\text{p = 1 }}\end{array}$

In order to figure out the reaction equation the order of the reactants needed, which is calculated by comparing any two experiments where the concentration of two reactants are constant and another varies, and in vice-versa. Hence, Rate equation is

$\text{Reaction rate = k }{\left[\text{A}\right]}^2{\left[\text{B}\right]}^0{\left[\text{C}\right]}^1$.

The respective order of reactant (A) is SECOND ORDER.

The respective order of reactant (B) is ZERO ORDER.

The respective order of reactant (C) is FIRST ORDER.

Hence, the reaction rate becomes,

$\text{Reaction rate = k }{\left[\text{A}\right]}^2{\left[\text{C}\right]}^1$

(b)

Interpretation Introduction

Interpretation: The value of ‘k’ of the given reaction has to be found.

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

The value of rate constant:

$\text{Reaction rate = k }{\left[\text{A}\right]}^2{\left[\text{C}\right]}^1$

Consider, any one of the given experimental value,

$\begin{array}{l}\text{Experiment 1: }\\ \\ \left[\text{A}\right]\text{ = }0.024\text{ ; }\left[\text{C}\right]\text{ =}0.032;\text{Reaction}\text{rate = }6.0\times {10}^{-6}\end{array}$

Substituting the above values into the reaction rate, the rate constant ‘k’ value is obtained as follows,

$\begin{array}{l}\text{Reaction rate = k }{\left[\text{A}\right]}^2{\left[\text{C}\right]}^1\\ \\ \text{6}{\text{.0×10}}^{\text{-6}}\text{ mol/L}\text{.s = k }{\left(0.024\right)}^2{\left(0.032\right)}^1\\ \\ \text{k = }\frac{\text{6}{\text{.0×10}}^{\text{-6}}\text{ mol/L}\text{.s }}{{\left(0.024\text{mol/L}\right)}^2{\left(0.032\text{mol/L}\right)}^1}\\ \\ \boxed{\text{k}\text{=}0.33L^2/mo{l}^2.s}.\end{array}$

The rate constant value is obtained as shown above. By substituting the any one of the concentrations of reactants and the initial rate into the reaction equation obtained at first.

Hence, the value of rate constant is $0.33L^2/mo{l}^2.s.$

(c)

Interpretation Introduction

Interpretation: The rate law for the given reaction has to be written.

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

The value of rate constant:

$\text{Reaction rate = k }{\left[\text{A}\right]}^2{\left[\text{C}\right]}^1$

Consider, any one of the given experimental value,

$\begin{array}{l}\text{Experiment 1: }\\ \\ \left[\text{A}\right]\text{ = }0.024\text{ ; }\left[\text{C}\right]\text{ =}0.032;\text{Reaction}\text{rate = }6.0\times {10}^{-6}\end{array}$

Substituting the above values into the reaction rate, the rate constant ‘k’ value is obtained as follows,

$\begin{array}{l}\text{Reaction rate = k }{\left[\text{A}\right]}^2{\left[\text{C}\right]}^1\\ \\ \text{6}{\text{.0×10}}^{\text{-6}}\text{ mol/L}\text{.s = k }{\left(0.024\right)}^2{\left(0.032\right)}^1\\ \\ \text{k = }\frac{\text{6}{\text{.0×10}}^{\text{-6}}\text{ mol/L}\text{.s }}{{\left(0.024\text{mol/L}\right)}^2{\left(0.032\text{mol/L}\right)}^1}\\ \\ \boxed{\text{k}\text{=}0.33L^2/mo{l}^2.s}.\end{array}$

The rate constant value is obtained as shown above. By substituting the any one of the concentrations of reactants and the initial rate into the reaction equation obtained at first.

Hence, the value of rate constant is $\text{0}{\text{.33 L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{.s}\text{.}$

By substituting the rate constant value into the obtained rate law,

The rate law becomes, $\text{Reaction rate = }\left(0.33L^2/mo{l}^2.s\right){\left[A\right]}^2{\left[C\right]}^1.$

(d)

Interpretation Introduction

Interpretation:

The rate in terms of changes with time for each of the components has to be written.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

For a general reaction,

$aA+bB\to cC+dD$ , Here a, b, c and d are the coefficients.

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}.$

The negative sign indicates the reduction of concentration of reactant.

$\Delta \left[A\right],\Delta \left[B\right],\Delta \left[C\right]and\Delta \left[D\right]$ are the change in concentration of A, B, C and D over time period $\Delta t$

Explanation of Solution

The balanced equation for given reaction is,

$\text{Given}\text{reaction: 2A + 2B}+\text{C}\to \text{D + 3E}$

Rate expression of the above reaction is,

$\text{Rate}\text{=}-\frac{1}{2}\frac{\text{Δ}\left[A\right]}{\text{Δt}}\text{=}-\frac{1}{2}\frac{\text{Δ}\left[B\right]}{\text{Δt}}=-\frac{\text{Δ}\left[C\right]}{\text{Δt}}=\frac{\text{Δ}\left[D\right]}{\text{Δt}}=\frac{1}{3}\frac{\text{Δ}\left[E\right]}{\text{Δt}}.$

The negative sign indicates the reduction of concentration of reactant A, B, and C. The coefficient of reactant A, B, and product E are written in reciprocal as $\frac{1}{2}$, $\frac{1}{2}$ and $\frac{1}{3}$, respectively as shown above.