Chapter 15, Problem 22E

(a)

Interpretation Introduction

Interpretation:

The order of the reaction with respect to Hb and CO should be predicted.

Concept Introduction:

Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduce amount of reactant in per unit time or increase amount of product in per unit time.

Answer to Problem 22E

The order of the reaction with respect to Hb and CO is 1^st for both.

Explanation of Solution

The initial concentration and rate constant is given as follows:

  

The rate law for the given reaction is represented as follows:

  $\text{Rate}=k{\left[\text{Hb}\right]}^a{\left[\text{CO}\right]}^b$ ...(1)

Where,

• $k$ is the rate constant.
• a and b is the order of the reaction with respect to $\text{Hb}$ and $\text{CO}$.

Substitute the values of 1^st row and 2^nd in equation (1) and divide 1^st row and 2^nd row values as shown below:

  $\begin{array}{c}\frac{{\left(\text{Rate}\right)}_{\text{1}}}{{\left(\text{Rate}\right)}_{\text{2}}}=\frac{{\left(k{\left[\text{Hb}\right]}^a{\left[\text{CO}\right]}^b\right)}_1}{{\left(k{\left[\text{Hb}\right]}^a{\left[\text{CO}\right]}^b\right)}_2}\\ \frac{0.619}{1.24}=\frac{\cancel{k}{\left(2.21\right)}^a\cancel{{\left(1.00\right)}^b}}{\cancel{k}{\left(4.42\right)}^a\cancel{{\left(1.00\right)}^b}}\\ 0.5={\left(0.5\right)}^a\\ a=1\end{array}$

Therefore, the order of reaction with respect to $\text{Hb}$ is 1.

Similarly, substitute the values of 2^nd and 3^rd row in equation (1) and divide 2^nd by 3^rd values as shown below:

  $\begin{array}{c}\frac{{\left(\text{Rate}\right)}_{\text{2}}}{{\left(\text{Rate}\right)}_{\text{3}}}=\frac{{\left(k{\left[\text{Hb}\right]}^a{\left[\text{CO}\right]}^b\right)}_2}{{\left(k{\left[\text{Hb}\right]}^a{\left[\text{CO}\right]}^b\right)}_3}\\ \frac{1.24}{3.71}=\frac{\cancel{k}\cancel{{\left(4.42\right)}^a}{\left(1.00\right)}^b}{\cancel{k}\cancel{{\left(4.42\right)}^a}{\left(3.00\right)}^b}\\ 0.33={\left(0.33\right)}^b\\ b=1\end{array}$

Therefore, the order of reaction with respect to $\text{CO}$ is 1.

(b)

Interpretation Introduction

Interpretation:

The rate law for the reaction should be stated.

Concept Introduction:

Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduce amount of reactant in per unit time or increase amount of product in per unit time.

Answer to Problem 22E

The rate law for the reaction is $k\left[\text{Hb}\right]\left[\text{CO}\right]$.

Explanation of Solution

The rate law for the given reaction is represented as follows:

  $\text{Rate}=k{\left[\text{Hb}\right]}^a{\left[\text{CO}\right]}^b$

Substitute a = 1 and b = 1 in the above expression.

  $\text{Rate}=k\left[\text{Hb}\right]\left[\text{CO}\right]$

Therefore, the rate law for the reaction is $k\left[\text{Hb}\right]\left[\text{CO}\right]$.

(c)

Interpretation Introduction

Interpretation:

The value of rate constant for the reaction should be predicted.

Concept Introduction:

Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduce amount of reactant in per unit time or increase amount of product in per unit time.

Answer to Problem 22E

The value of rate constant for the given reaction is $0.280\mu {\text{mol}}^{-1}\text{L}{\text{s}}^{-1}$.

Explanation of Solution

The rate law for the given reaction is represented as follows:

  $\text{Rate}=k\left[\text{Hb}\right]\left[\text{CO}\right]$

Substitute 1^st row values of given table in the above expression.

  $\begin{array}{c}0.619\mu \text{mol}{\text{L}}^{-1}{\text{s}}^{-1}=k\left(2.21\mu \text{mol/L}\right)\left(1.00\mu \text{mol/L}\right)\\ k=\frac{0.619\mu \text{mol}{\text{L}}^{-1}{\text{s}}^{-1}}{\left(2.21\mu \text{mol/L}\right)\left(1.00\mu \text{mol/L}\right)}\\ =0.280\mu {\text{mol}}^{-1}\text{L}{\text{s}}^{-1}\end{array}$

Therefore, the value of rate constant for the given reaction is $0.280\mu {\text{mol}}^{-1}\text{L}{\text{s}}^{-1}$.

(d)

Interpretation Introduction

Interpretation:

The initial rate for an experiment where ${\left[\text{Hb}\right]}_0=3.36\mu \text{mol/L}$ and ${\left[\text{CO}\right]}_0=2.40\mu \text{mol/L}$ should be calculated.

Concept Introduction:

Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduce amount of reactant in per unit time or increase amount of product in per unit time.

Answer to Problem 22E

The initial rate for an experiment where ${\left[\text{Hb}\right]}_0=3.36\mu \text{mol/L}$ and ${\left[\text{CO}\right]}_0=2.40\mu \text{mol/L}$ is $2.26\mu \text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$.

Explanation of Solution

The rate law for the given reaction is represented as follows:

  $\text{Rate}=k\left[\text{Hb}\right]\left[\text{CO}\right]$

Substitute the value of ${\left[\text{Hb}\right]}_0=3.36\mu \text{mol/L}$ , ${\left[\text{CO}\right]}_0=2.40\mu \text{mol/L}$ and $k=0.280\mu {\text{mol}}^{-1}\text{L}{\text{s}}^{-1}$ in the above expression.

  $\begin{array}{c}\text{rate}=0.280\mu {\text{mol}}^{-1}\text{L}{\text{s}}^{-1}\left(3.36\mu \text{mol/L}\right)\left(2.40\mu \text{mol/L}\right)\\ =2.26\mu \text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

Therefore, the initial rate for an experiment where ${\left[\text{Hb}\right]}_0=3.36\mu \text{mol/L}$ and ${\left[\text{CO}\right]}_0=2.40\mu \text{mol/L}$ is $2.26\mu \text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$.