Chapter 17, Problem 6ALQ

Interpretation Introduction

Interpretation:

The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as,

$\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons \text{c}\text{C}+\text{d}\text{D}$

The equilibrium constant for the above chemical reaction is expressed as,

$\text{K}=\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$.

Answer to Problem 6ALQ

The equilibrium constant for the given chemical reaction is expressed as,

$\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

At equilibrium $\left[\text{A}\right]=2\text{M},\left[\text{B}\right]=1\text{M}$ and $\left[\text{C}\right]=4\text{M}$.

Substitute the value of equilibrium concentration in the above equation.

$\begin{array}{c}\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\\ =\frac{\left[\text{4}\right]}{\left[\text{2}\right]\left[\text{1}\right]}\\ =\text{ 2}\end{array}$

The equilibrium constant for the given chemical reaction is $2$.

Hence, the first statement is correct.

It is given that after adding the $\text{3 moles}$ of B, the possible equilibrium condition is $\left[\text{A}\right]=1\text{M},\left[\text{B}\right]=3\text{M}$ and $\left[\text{C}\right]=6\text{M}$.

But after the addition of $\text{3 moles}$ of B, the concentration of B will increase to $4\text{M}$ and the concentration of the A will be same as $2\text{M}$. Hence, the value of equilibrium constant will be change.

The correct statements are given below.

The possible equilibrium condition will be $\left[\text{A}\right]=2\text{M},\left[\text{B}\right]=4\text{M}$ and $\left[\text{C}\right]=6\text{M}$.

The equilibrium constant for the given chemical reaction is calculated as follows:

$\begin{array}{c}\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\\ =\frac{\left[6\right]}{\left[4\right]\left[2\right]}\\ =\text{ 0}\text{.75}\end{array}$

The new equilibrium constant for the given chemical reaction will be $\text{ 0}\text{.75}$.

Explanation of Solution

The given reaction is,

$\text{A}(g)+\text{B}(g)\rightleftharpoons \text{C}(g)$

At equilibrium $\left[\text{A}\right]=2\text{M},\left[\text{B}\right]=1\text{M}$ and $\left[\text{C}\right]=4\text{M}$.

The equilibrium constant for the given chemical reaction is expressed as,

$\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

Substitute the value of equilibrium concentration in the above equation.

$\begin{array}{c}\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\\ =\frac{\left[\text{4}\right]}{\left[\text{2}\right]\left[\text{1}\right]}\\ =\text{ 2}\end{array}$

The equilibrium constant for the given chemical reaction is $2$.

Hence, the first statement is correct.

It is given that after adding the $\text{3 moles}$ of B, the possible equilibrium condition is $\left[\text{A}\right]=1\text{M},\left[\text{B}\right]=3\text{M}$ and $\left[\text{C}\right]=6\text{M}$.

But after the addition of $\text{3 moles}$ of B to the first statement, the concentration of B will increase to $4\text{M}$ and the concentration of the A will be same as $2\text{M}$. Hence, the value of equilibrium constant will be change.

Therefore, the possible equilibrium condition will be $\left[\text{A}\right]=2\text{M},\left[\text{B}\right]=4\text{M}$ and $\left[\text{C}\right]=6\text{M}$.

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is expressed as,

$\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

Substitute the value of equilibrium concentration in the above equation

$\begin{array}{c}\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\\ =\frac{\left[6\right]}{\left[4\right]\left[2\right]}\\ =\text{ 0}\text{.75}\end{array}$

The new equilibrium constant for the given chemical reaction will be $\text{ 0}\text{.75}$.

Hence, the first statement is correct and second statement is incorrect.

Conclusion

The equilibrium constant for the given chemical reaction is expressed as,

$\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

At equilibrium $\left[\text{A}\right]=2\text{M},\left[\text{B}\right]=1\text{M}$ and $\left[\text{C}\right]=4\text{M}$.

Put the value of equilibrium concentration in the above equation.

$\begin{array}{c}\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\\ =\frac{\left[\text{4}\right]}{\left[\text{2}\right]\left[\text{1}\right]}\\ =\text{ 2}\end{array}$

The equilibrium constant for the given chemical reaction is $2$.

Hence, the first statement is correct.

It is given that after adding the $\text{3 moles}$ of B, the possible equilibrium condition is $\left[\text{A}\right]=1\text{M},\left[\text{B}\right]=3\text{M}$ and $\left[\text{C}\right]=6\text{M}$.

But after the addition of $\text{3 moles}$ of B, the concentration of B will increase to $4\text{M}$ and the concentration of the A will be same as $2\text{M}$. Hence, the value of equilibrium constant will be change.

The possible equilibrium condition will be $\left[\text{A}\right]=2\text{M},\left[\text{B}\right]=4\text{M}$ and $\left[\text{C}\right]=6\text{M}$.

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is calculated as follows:

$\begin{array}{c}\text{K}=\frac{\left[\text{C}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\\ =\frac{\left[6\right]}{\left[4\right]\left[2\right]}\\ =\text{ 0}\text{.75}\end{array}$

The new equilibrium constant for the given chemical reaction will be $\text{ 0}\text{.75}$.