Chapter 7, Problem 7A.1AST

Interpretation Introduction

Interpretation:

Theaverage rate of the reaction has to be determined.

Concept Introduction:

The average rate of the reaction is referred to the change in the molar concentration in the distinct interval of time.  Mathematically, the ratio of change in molar concentration of a reactant or a product to the change in the time interval gives the average rate of reaction.

The average rate of consumption or the disappearance of reactant is negative.  Mathematical expression is shown below.

  $\text{Average rate of consumption of R }=-\frac{\Delta \left[\text{R}\right]}{\Delta t}$

The average rate of formation or the appearance of reactant is positive.  Mathematical expression is shown below.

  $\text{Average rate of formation of P }=\frac{\Delta \left[\text{P}\right]}{\Delta t}$

Explanation of Solution

The given chemical equation for the formation of hydrogen iodide is shown below.

  ${\text{H}}_2\left(\text{g}\right)+{\text{I}}_2\left(\text{g}\right)\to 2\text{HI}\left(\text{g}\right)$

As per the given data in the question, the molar concentration of hydrogen iodide is increased from $4.20\left(\text{mmol}\text{HI}\right)\cdot {\text{L}}^{-1}$ to $6.00\left(\text{mmol}\text{HI}\right)\cdot {\text{L}}^{-1}$ on increasing the temperature.

The change in the concentration of hydrogen iodide is calculated as shown below.

  $\begin{array}{c}\Delta \left[\text{P}\right]=6.00\left(\text{mmol}\text{HI}\right)\cdot {\text{L}}^{-1}-4.20\left(\text{mmol}\text{HI}\right)\cdot {\text{L}}^{-1}\\ =1.80\left(\text{mmol}\text{HI}\right)\cdot {\text{L}}^{-1}\end{array}$

The relation to calculate the average rate of the reaction is shown below.

  $\text{Average rate of formation of HI}=\frac{\Delta \left[\text{P}\right]}{\Delta t}$        (1)

Where,

• $\Delta \left[\text{P}\right]$ is the change in molar concentration of $\text{HI}$.
• $\Delta t$ is the change in time.

The value of $\Delta \left[\text{P}\right]$ is $1.80\left(\text{mmol}\text{HI}\right)\cdot {\text{L}}^{-1}$.

The value of $\Delta t$ is $200\text{s}$.

Substitute the value of $\Delta \left[\text{P}\right]$ and $\Delta t$ in equation (1).

  $\begin{array}{c}\text{Average rate of formation of HI }=\frac{1.80\left(\text{mmol}\text{HI}\right)\cdot {\text{L}}^{-1}}{200\text{s}}\\ =9.00\times {10}^{-3}\left(\text{mmol}\text{HI}\right)\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}\times \frac{{10}^3\text{μmol}}{1\text{mmol}}\\ =9.00\left(\text{μmol}\text{HI}\right)\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}\end{array}$

Thus, the average rate for the formation of hydrogen iodide is $\underset{\_}{9.00\left(\mu molHI\right)\cdot L^{-1}\cdot s^{-1}}$.