Chapter 18, Problem 108A

(a)

Interpretation Introduction

Interpretation: The graph of concentration (y-axis) versus time (x-axis) has to be sketched.

Concept Introduction: The rate is a method for determining how much something changes over a predetermined period of time. The amount of reactant or product that changes per unit of time is the reaction rate.

Answer to Problem 108A

The graph of concentration (y-axis) versus time (x-axis) is as follows:

  

Explanation of Solution

The change in concentration of a substance with respect to unit time gives the rate of a reaction. The mathematical equation for the rate is given as follows:

  $\text{Rate = ±}\frac{\text{ΔA}}{\text{Δt}}$   ...... (1)

The rate of the reaction is monitored by either the appearance of the product or the disappearance of the reactant over time.

  

(b)

Interpretation Introduction

Interpretation: The rate of the reaction at $\text{t =100}\text{seconds}$ and $\text{t =250}\text{seconds}$ are to be determined.

Concept Introduction: The rate is a method for determining how much something changes over a predetermined period of time. The amount of reactant or product that changes per unit of time is the reaction rate. The change in concentration of a substance with respect to unit time gives the rate of a reaction. The mathematical equation for the rate is given as follows:

  $\text{Rate = ±}\frac{\text{ΔA}}{\text{Δt}}$   ...... (1)

Here, $\Delta \text{A=A}-{\text{A}}_0$ which is the concentration and $\text{Δt}$ is the time taken for the reaction.

Answer to Problem 108A

The rate of the reaction at $\text{t =100}\text{seconds}$ and $\text{t =250}\text{seconds}$ are $9.9\times {10}^{-4}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ and $7.56\times {10}^{-4}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ .

Explanation of Solution

Given information:

The initial concentration is ${\text{A}}_{\text{0}}=0.3\text{mol}{\text{L}}^{-1}$ .

The given time is $\text{t =100}\text{seconds}$ and $\text{t =250}\text{seconds}$ .

The concentration at $\text{t =100}\text{seconds}$ is $\text{A}=0.201\text{mol}{\text{L}}^{-1}$ .

The concentration at $\text{t =250}\text{seconds}$ is $\text{A}=0.111\text{mol}{\text{L}}^{-1}$ .

The change in concentration of a substance with respect to unit time gives the rate of a reaction. The mathematical equation for the rate is given as follows:

  $\text{Rate = ±}\frac{\text{ΔA}}{\text{Δt}}$   ...... (1)

Here, $\Delta \text{A=A}-{\text{A}}_0$ which is the concentration and $\text{Δt}$ is the time taken for the reaction.

The rate of the reaction is monitored by either the appearance of the product or the disappearance of the reactant over time.

To calculate the rate at $\text{t =100}\text{seconds}$ , substitute the values in equation (1).

  $\begin{array}{c}\text{Rate = ±}\frac{\text{A}-{\text{A}}_0}{\text{Δt}}\\ =-\frac{0.201\text{mol}{\text{L}}^{-1}-0.3\text{mol}{\text{L}}^{-1}}{100\text{s}}\\ =\frac{0.099}{100}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\\ =9.9\times {10}^{-4}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

To calculate the rate at $\text{t =250}\text{seconds}$ , substitute the values in equation (1).

  $\begin{array}{c}\text{Rate = ±}\frac{\text{A}-{\text{A}}_0}{\text{Δt}}\\ =-\frac{0.111\text{mol}{\text{L}}^{-1}-0.3\text{mol}{\text{L}}^{-1}}{250\text{s}}\\ =\frac{0.189}{250}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\\ =7.56\times {10}^{-4}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$