Chapter 15, Problem 40GQ

Consider the following equilibrium:

COBr₂(g) ⇄ CO(g) + Br₂(g)    K_c = 0.190 at 73 °C

1. (a) A 0.50 mol sample of COBr₂ is transferred to a 9.50-L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.
2. (b) The volume of the container is decreased to 4.5 L and the system allowed to return to equilibrium. Calculate the new equilibrium concentrations. (Hint: The calculation will be easier if you view this as a new problem with 0.5 mol of COBr₂ transferred to a 4.5-L flask.)
3. (c) What is the effect of decreasing the container volume from 9.50 L to 4.50 L?

Interpretation Introduction

Interpretation:

The equilibrium concentration of each species in the reaction ${\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}$ in a $\text{9}\text{.50}\text{L}$ flask and the new equilibrium concentration when the volume changes to $\text{4}\text{.5}\text{L}$ and the effect of decreasing volume is to be given.

Concept Introduction:

Equilibrium constant in terms of concentration${\text{[K}}_{\text{C}}\text{]}$: Equilibrium constant can be expressed in terms of concentration.

$\begin{array}{l}\text{aA}{}_{\text{(g)}}\text{+}{\text{bB}}_{\text{(g)}}\rightleftarrows \text{cC}{}_{\text{(g)}}\text{+}\text{dD}{}_{\text{(g)}}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}\text{×}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}\text{×}{\left[\text{B}\right]}^{\text{b}}}\end{array}$

$\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{(L)}}$

Le Chatelier’s principle: If an equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

Factor’s that effect chemical equilibria:

Concentration – Equilibrium will be affected by changing the concentration of reactant or product. If we increase the concentration of reactant system will try to reverse the change by favouring forward reaction and thus increase the concentration of products. Like wise adding products increase yield of reactants.

Temperature – When the temperature increases equilibrium will shift in the endothermic direction, in the direction that absorbs heat. When the temperature decreases equilibrium will shift in the exothermic direction, in the direction that releases heat.

Pressure – If the reaction consists of only liquid and solid reactants and products, pressure has no effect in the equilibrium.

In gas reactions if the number of moles has no change then there will be no effect by pressure on equilibrium.

If pressure increases (volume decreases) then equilibrium will shift to the direction having less number of molecules and if pressure decreases (volume increases) system will shift to the direction having more number of molecules.

Answer to Problem 40GQ

The equilibrium concentration of each species when the volume is $\text{9}\text{.50}\text{L}$ is

$\left[{\text{COBr}}_{\text{2}}\right]=\text{9}{\text{.7×10}}^{\text{-3}}\text{M}$, $\left[\text{CO}\right]\text{=}\left[{\text{Br}}_{\text{2}}\right]\text{=}\text{x}\text{=}\text{0}\text{.0429}\text{M}$

Explanation of Solution

To determine:

The equilibrium concentration of each species in the reaction ${\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}$ when the volume is $\text{9}\text{.50}\text{L}$.

Given:

$\begin{array}{c}{\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}\\ {\text{K}}_{\text{c}}\text{=}\text{0}\text{.190}\\ \text{T}\text{=}\text{73°C }\text{=}\text{346}\text{K}\\ \text{Number}\text{of}\text{moles}\text{of}{\text{COBr}}_{\text{2}}\text{=}\text{0}\text{.50}\text{mol}\\ \text{V}\text{=}\text{9}\text{.50}\text{L}\end{array}$

$\begin{array}{l}\text{Concentration}\text{of}{\text{COBr}}_{\text{2}}\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}}\\ \text{=}\frac{\text{0}\text{.50}}{\text{9}\text{.50}}\\ \text{=}\text{0}\text{.0526}\text{M}\end{array}$

$\begin{array}{l}{\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}\\ \text{Initial}\text{0}\text{.0526}\text{0}\text{0}\\ \text{Equilibrium}\text{0}\text{.0526-x}\text{x}\text{x}\end{array}$

$\begin{array}{c}{\text{K}}_{\text{c}}\text{=}\frac{\left[\text{CO}\right]\left[{\text{Br}}_{\text{2}}\right]}{\left[{\text{COBr}}_{\text{2}}\right]}\\ \text{0}\text{.190}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.0526-x}}\\ {\text{x}}^{\text{2}}\text{+0}\text{.190x-9}{\text{.994×10}}^{\text{-3}}\text{=}\text{0}\\ \text{Solving}\text{this}\text{we}\text{get}\text{x}\text{=}\text{0}\text{.0429}\end{array}$

$\begin{array}{l}\left[{\text{COBr}}_{\text{2}}\right]\text{=}\text{0}\text{.0526}\text{-}\text{x}\\ \text{=}\text{0}\text{.0526-0}\text{.0429}\\ \text{=}\text{9}{\text{.7×10}}^{\text{-3}}\text{M}\end{array}$

$\left[\text{CO}\right]\text{=}\left[{\text{Br}}_{\text{2}}\right]\text{=}\text{x}\text{=}\text{0}\text{.0429}\text{M}$

Interpretation Introduction

Interpretation:

The equilibrium concentration of each species in the reaction ${\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}$ in a $\text{9}\text{.50}\text{L}$ flask and the new equilibrium concentration when the volume changes to $\text{4}\text{.5}\text{L}$ and the effect of decreasing volume is to be given.

Concept Introduction:

Equilibrium constant in terms of concentration${\text{[K}}_{\text{C}}\text{]}$: Equilibrium constant can be expressed in terms of concentration.

$\begin{array}{l}\text{aA}{}_{\text{(g)}}\text{+}{\text{bB}}_{\text{(g)}}\rightleftarrows \text{cC}{}_{\text{(g)}}\text{+}\text{dD}{}_{\text{(g)}}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}\text{×}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}\text{×}{\left[\text{B}\right]}^{\text{b}}}\end{array}$

$\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{(L)}}$

Le Chatelier’s principle: If an equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

Factor’s that effect chemical equilibria:

Concentration – Equilibrium will be affected by changing the concentration of reactant or product. If we increase the concentration of reactant system will try to reverse the change by favouring forward reaction and thus increase the concentration of products. Like wise adding products increase yield of reactants.

Temperature – When the temperature increases equilibrium will shift in the endothermic direction, in the direction that absorbs heat. When the temperature decreases equilibrium will shift in the exothermic direction, in the direction that releases heat.

Pressure – If the reaction consists of only liquid and solid reactants and products, pressure has no effect in the equilibrium.

In gas reactions if the number of moles has no change then there will be no effect by pressure on equilibrium.

If pressure increases (volume decreases) then equilibrium will shift to the direction having less number of molecules and if pressure decreases (volume increases) system will shift to the direction having more number of molecules.

Answer to Problem 40GQ

The equilibrium concentration of each species when the volume is $\text{4}\text{.5}\text{L}$ is

$\left[{\text{COBr}}_{\text{2}}\right]\text{=}0.0325\text{M}$, $\left[\text{CO}\right]\text{=}\left[{\text{Br}}_{\text{2}}\right]\text{=}\text{x}\text{=}\text{0}\text{.0785}\text{M}$

Explanation of Solution

To determine:

The equilibrium concentration of each species in the reaction ${\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}$ when the volume is $\text{4}\text{.50}\text{L}$.

Given:

$\begin{array}{c}{\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}\\ {\text{K}}_{\text{c}}\text{=}\text{0}\text{.190}\\ \text{T}\text{=}\text{73°C }\text{=}\text{346}\text{K}\\ \text{Number}\text{of}\text{moles}\text{of}{\text{COBr}}_{\text{2}}\text{=}\text{0}\text{.50}\text{mol}\\ \text{V}\text{=}\text{4}\text{.50}\text{L}\end{array}$

$\begin{array}{l}\text{Concentration}\text{of}{\text{COBr}}_{\text{2}}\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}}\\ \text{=}\frac{\text{0}\text{.50}}{\text{4}\text{.50}}\\ \text{=}\text{0}\text{.111}\text{M}\end{array}$

$\begin{array}{l}{\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}\\ \text{Initial}\text{0}\text{.111}\text{0}\text{0}\\ \text{Equilibrium}\text{0}\text{.111-x}\text{x}\text{x}\end{array}$

$\begin{array}{c}{\text{K}}_{\text{c}}\text{=}\frac{\left[\text{CO}\right]\left[{\text{Br}}_{\text{2}}\right]}{\left[{\text{COBr}}_{\text{2}}\right]}\\ \text{0}\text{.190}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.111-x}}\\ {\text{x}}^{\text{2}}\text{+0}\text{.190x-0}\text{.02109}\text{=}\text{0}\\ \text{Solving}\text{this}\text{we}\text{get}\text{x}\text{=}\text{0}\text{.0785}\end{array}$

$\begin{array}{l}\left[{\text{COBr}}_{\text{2}}\right]\text{=}\text{0}\text{.111}\text{-}\text{x}\\ \text{=}\text{0}\text{.111-0}\text{.0785}\\ \text{=}0.0325\text{M}\end{array}$

$\left[\text{CO}\right]\text{=}\left[{\text{Br}}_{\text{2}}\right]\text{=}\text{x}\text{=}\text{0}\text{.0785}\text{M}$

Interpretation Introduction

Interpretation:

The equilibrium concentration of each species in the reaction ${\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}$ in a $\text{9}\text{.50}\text{L}$ flask and the new equilibrium concentration when the volume changes to $\text{4}\text{.5}\text{L}$ and the effect of decreasing volume is to be given.

Concept Introduction:

Equilibrium constant in terms of concentration${\text{[K}}_{\text{C}}\text{]}$: Equilibrium constant can be expressed in terms of concentration.

$\begin{array}{l}\text{aA}{}_{\text{(g)}}\text{+}{\text{bB}}_{\text{(g)}}\rightleftarrows \text{cC}{}_{\text{(g)}}\text{+}\text{dD}{}_{\text{(g)}}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}\text{×}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}\text{×}{\left[\text{B}\right]}^{\text{b}}}\end{array}$

$\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{(L)}}$

Le Chatelier’s principle: If an equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

Factor’s that effect chemical equilibria:

Concentration – Equilibrium will be affected by changing the concentration of reactant or product. If we increase the concentration of reactant system will try to reverse the change by favouring forward reaction and thus increase the concentration of products. Like wise adding products increase yield of reactants.

Temperature – When the temperature increases equilibrium will shift in the endothermic direction, in the direction that absorbs heat. When the temperature decreases equilibrium will shift in the exothermic direction, in the direction that releases heat.

Pressure – If the reaction consists of only liquid and solid reactants and products, pressure has no effect in the equilibrium.

In gas reactions if the number of moles has no change then there will be no effect by pressure on equilibrium.

If pressure increases (volume decreases) then equilibrium will shift to the direction having less number of molecules and if pressure decreases (volume increases) system will shift to the direction having more number of molecules.

Answer to Problem 40GQ

The effect of decreasing volume is explained according to Le Chatelier’s principle.

Explanation of Solution

To determine:

The effect of decreasing volume in equilibrium

The equilibrium concentrations of the species with different volumes are calculated.

When the volume reduces from $\text{9}\text{.50}\text{L}$ to $\text{4}\text{.50}\text{L}$, the equilibrium concentration of ${\text{COBr}}_{\text{2}}$ changes from $\text{9}{\text{.7×10}}^{\text{-3}}\text{M}$ to $0.0325\text{M}$. This can be explained using Le Chatelier’s principle.

According to Le Chatelier’s principle ,if an equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

In gas reactions if the number of moles has no change then there will be no effect by pressure on equilibrium.

If volume decreases then equilibrium will shift to the direction having less number of molecules and if volume increases system will shift to the direction having more number of molecules.

Here volume is reduced and as a result, equilibrium will shift in the direction that has less number of moles.

${\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}$

Therefore, for the above reaction equilibrium will shift to left side and thus concentration of ${\text{COBr}}_{\text{2}}$ increases when volume changes from $\text{9}\text{.50}\text{L}$ to $\text{4}\text{.50}\text{L}$.

Conclusion:

The equilibrium concentration of each species in the reaction ${\text{COBr}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{CO}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}\text{(g)}}$ in a $\text{9}\text{.50}\text{L}$ flask and the new equilibrium concentration when the volume changes to $\text{4}\text{.5}\text{L}$ and the effect of decreasing volume is calculated and explained.