Chapter 12, Problem 37E

The rote of the reaction between hemoglobin (Hb) and carbon monoxide (CO) was studied at 20°C. The following data were collected with all concentration units in μmol/L. (A hemoglobin concentration of 2.21 μmol/L is equal to 2.21 × 10⁻⁶ mol/L.)

[Hb]0(μmol/L)	[CO]0(μmol/L)	Initial Rate (μmol/L · s)
2.21	1.00	0.619
4.42	1.00	1.24
4.42	3.00	3.71

a. Determine the orders of this reaction with respect to Hb and CO.

b. Determine the rate law.

c. Calculate the value of the rate constant.

d. What would be the initial rate for an experiment with [Hb]₀ = 3.36 μmol/L and [CO]₀ = 2.40 μmol/L?

Interpretation Introduction

Interpretation: The data related to the rate of reaction at different initial concentrations of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ is given for the reaction between hemoglobin $\left(\text{Hb}\right)$ and carbon monoxide $\left(\text{CO}\right)$. The orders of the reaction with respect to $\text{Hb}$ and $\text{CO}$, rate law, the value of the rate constant and the initial rate of the reaction for the given values of concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

Answer to Problem 37E

The order of the reaction with respect to $\text{Hb}$ is $\underset{\_}{1}$ and with respect to $\text{CO}$ is also $\underset{\_}{1}$.

Explanation of Solution

The rate law for the given reaction is calculated by the expression,

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

Where,

• $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ are the concentrations of $\text{Hb}$ and $\text{CO}$.
• $k$ is the rate constant.

The values of $m$ and $n$ are calculated by the comparison of the different rate values from the given table.

The value of $m$ is calculated using the first and second result. Substitute the values of the concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ for the first two experiments in the above expression.

$\text{Rate}\text{1}=k{\left[2.21\right]}^m{\left[1.00\right]}^n$

$\text{Rate}2=k{\left[4.42\right]}^m{\left[1.00\right]}^n$

According to the given rate values in the table,

$\frac{\text{Rate}\text{2}}{\text{Rate}\text{1}}=\frac{1.24\text{μmol}/\text{L}\cdot \text{s}}{0.619\text{μmol}/\text{L}\cdot \text{s}}$

Therefore,

$\begin{array}{c}\frac{k{\left[4.42\right]}^m{\left[1.00\right]}^n}{k{\left[2.21\right]}^m{\left[1.00\right]}^n}=\frac{1.24\text{μmol}/\text{L}\cdot \text{s}}{0.619\text{μmol}/\text{L}\cdot \text{s}}\\ {\left(2\right)}^m=2\\ m=\underset{\_}{1}\end{array}$

The value of $n$ is calculated using the second and third result. Substitute the values of the concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ for the second and third experiments in the above expression.

$\text{Rate}2=k{\left[4.42\right]}^m{\left[1.00\right]}^n$

$\text{Rate}3=k{\left[4.42\right]}^m{\left[3.00\right]}^n$

According to the given rate values in the table,

$\frac{\text{Rate}3}{\text{Rate}2}=\frac{3.71\text{μ}\text{mol}/\text{L}\cdot \text{s}}{1.24\text{μmol}/\text{L}\cdot \text{s}}$

Therefore,

$\begin{array}{c}\frac{k{\left[4.42\right]}^m{\left[3.00\right]}^n}{k{\left[4.42\right]}^m{\left[1.00\right]}^n}=\frac{3.71\text{μ}\text{mol}/\text{L}\cdot \text{s}}{1.24\text{μmol}/\text{L}\cdot \text{s}}\\ {\left(3\right)}^n=3\\ n=\underset{\_}{1}\end{array}$

Substitute the values of $m$ and $n$ in the rate law expression.

$\text{Rate}=k{\left[\text{Hb}\right]}^1{\left[\text{CO}\right]}^1$

Conclusion

The order of the reaction with respect to $\text{Hb}$ is $\underset{\_}{1}$ and with respect to $\text{CO}$ is also $\underset{\_}{1}$.

Interpretation Introduction

Interpretation: The data related to the rate of reaction at different initial concentrations of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ is given for the reaction between hemoglobin $\left(\text{Hb}\right)$ and carbon monoxide $\left(\text{CO}\right)$. The orders of the reaction with respect to $\text{Hb}$ and $\text{CO}$, rate law, the value of the rate constant and the initial rate of the reaction for the given values of concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

Answer to Problem 37E

The rate law is, $\text{Rate}=k{\left[\text{Hb}\right]}^1{\left[\text{CO}\right]}^1$.

Explanation of Solution

The rate law for the given reaction is calculated by the expression,

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

Where,

• $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ are the concentrations of $\text{Hb}$ and $\text{CO}$.
• $k$ is the rate constant.

The values of $m$ and $n$ are calculated by the comparison of the different rate values from the given table.

The value of $m$ is calculated using the first and second result. Substitute the values of the concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ for the first two experiments in the above expression.

$\text{Rate}\text{1}=k{\left[2.21\right]}^m{\left[1.00\right]}^n$

$\text{Rate}2=k{\left[4.42\right]}^m{\left[1.00\right]}^n$

According to the given rate values in the table,

$\frac{\text{Rate}\text{2}}{\text{Rate}\text{1}}=\frac{1.24\text{μmol}/\text{L}\cdot \text{s}}{0.619\text{μmol}/\text{L}\cdot \text{s}}$

Therefore,

$\begin{array}{c}\frac{k{\left[4.42\right]}^m{\left[1.00\right]}^n}{k{\left[2.21\right]}^m{\left[1.00\right]}^n}=\frac{1.24\text{μmol}/\text{L}\cdot \text{s}}{0.619\text{μmol}/\text{L}\cdot \text{s}}\\ {\left(2\right)}^m=2\\ m=1\end{array}$

The value of $n$ is calculated using the second and third result. Substitute the values of the concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ for the second and third experiments in the above expression.

$\text{Rate}2=k{\left[4.42\right]}^m{\left[1.00\right]}^n$

$\text{Rate}3=k{\left[4.42\right]}^m{\left[3.00\right]}^n$

According to the given rate values in the table,

$\frac{\text{Rate}3}{\text{Rate}2}=\frac{3.71\text{μ}\text{mol}/\text{L}\cdot \text{s}}{1.24\text{μmol}/\text{L}\cdot \text{s}}$

Therefore,

$\begin{array}{c}\frac{k{\left[4.42\right]}^m{\left[3.00\right]}^n}{k{\left[4.42\right]}^m{\left[1.00\right]}^n}=\frac{3.71\text{μ}\text{mol}/\text{L}\cdot \text{s}}{1.24\text{μmol}/\text{L}\cdot \text{s}}\\ {\left(3\right)}^n=3\\ n=1\end{array}$

Substitute the values of $m$ and $n$ in the rate law expression.

$\text{Rate}=k{\left[\text{Hb}\right]}^1{\left[\text{CO}\right]}^1$

Conclusion

The rate law for the given reaction is $\text{Rate}=k{\left[\text{Hb}\right]}^1{\left[\text{CO}\right]}^1$.

Interpretation Introduction

Interpretation: The data related to the rate of reaction at different initial concentrations of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ is given for the reaction between hemoglobin $\left(\text{Hb}\right)$ and carbon monoxide $\left(\text{CO}\right)$. The orders of the reaction with respect to $\text{Hb}$ and $\text{CO}$, rate law, the value of the rate constant and the initial rate of the reaction for the given values of concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

To determine: The value of the rate constant, $k$.

Answer to Problem 37E

The value of $k$ is $\underset{\_}{0.28L/\mu mol\cdot s}$.

Explanation of Solution

The rate law for the given reaction is calculated by the expression,

$\text{Rate}=k{\left[\text{Hb}\right]}^1{\left[\text{CO}\right]}^1$

Substitute the given values of rate and the concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ in the above expression.

$\begin{array}{c}\text{Rate}=k{\left[\text{Hb}\right]}^1{\left[\text{CO}\right]}^1\\ 0.619\text{μmol}/\text{L}\cdot \text{s}=k{\left[\text{2}\text{.21}\text{μmol}/\text{L}\right]}^1{\left[1.00\text{μmol}/\text{L}\right]}^1\\ 0.619\text{μmol}/\text{L}\cdot \text{s}=k\left[2.21{{\text{μmol}}^2/\text{L}}^2\right]\end{array}$

Simplify the above expression.

$\begin{array}{c}k=\frac{0.619\text{μmol}/\text{L}\cdot \text{s}}{\left[2.21{{\text{μmol}}^2/\text{L}}^2\right]}\\ =\underset{\_}{0.28L/\mu mol\cdot s}\end{array}$

Conclusion

The value of $k$ for the given reaction is $\underset{\_}{0.28L/\mu mol\cdot s}$.

Interpretation Introduction

Interpretation: The data related to the rate of reaction at different initial concentrations of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ is given for the reaction between hemoglobin $\left(\text{Hb}\right)$ and carbon monoxide $\left(\text{CO}\right)$. The orders of the reaction with respect to $\text{Hb}$ and $\text{CO}$, rate law, the value of the rate constant and the initial rate of the reaction for the given values of concentration of $\left[\text{Hb}\right]$ and $\left[\text{CO}\right]$ are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

Answer to Problem 37E

Solution: The initial rate is $\underset{\_}{2.26\mu mol/L\cdot s}$.

Explanation of Solution

The rate law for the given reaction is calculated by the expression,

$\text{Rate}=k{\left[\text{Hb}\right]}^1{\left[\text{CO}\right]}^1$

The given values are,

$\begin{array}{l}{\left[\text{Hb}\right]}_{\text{ο}}=3.36\text{μmol}/\text{L}\\ {\left[\text{CO}\right]}_{\text{ο}}=2.40\text{μmol}/\text{L}\end{array}$

Substitute the value of rate and the given initial concentration of $\text{NO}$ and ${\text{O}}_2$ in the rate expression.

$\begin{array}{c}\text{Rate}=k{\left[\text{Hb}\right]}^1{\left[\text{CO}\right]}^1\\ =\left(0.280\text{L}/\text{μmol}\cdot \text{s}\right)\left[\left(3.36\right)\left(2.40\right)\text{μmol}/\text{L}\right]\\ =\underset{\_}{2.26\mu mol/L\cdot s}\end{array}$

Conclusion

The initial rate for the given initial concentration values of $\text{Hb}$ and $\text{CO}$. is $\underset{\_}{2.26\mu mol/L\cdot s}$.