Chapter 14, Problem 14.101QP

At high temperatures, a dynamic equilibrium exists between carbon monoxide, carbon dioxide, and solid carbon.

$\text{C}(s)+{\text{CO}}_2(g)\stackrel{}{\rightleftharpoons }2\text{CO(}g\text{)};\Delta H°=172.5\text{kJ}$

At 850°C, K_c is 0.153.

1. a What is the value of Kp?
2. b If the original reaction system consisted of just carbon and 1.50 atm of CO₂, what are the pressures of CO₂ and CO when equilibrium has been established?
3. c How will the equilibrium pressure of CO change if the temperature is decreased?

Interpretation Introduction

Interpretation:

The equilibrium constant has to be found and the equilibrium partial pressures of the reactant and product also has to be found.  The effect of decrease in temperature in the equilibrium constant has to be found.

Concept introduction:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where the reactant A is giving product B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Equilibrium constant $\left({\text{K}}_{\text{p}}\right)$: For gaseous substances the concentration will be proportional to its partial pressure.  The equilibrium constant for gaseous  reactions can be expressed in terms of the partial pressures of the gaseous products and reactants, and it is called equilibrium constant ${\text{K}}_{\text{p}}$  .  The value of ${\text{K}}_{\text{p}}$ is different from ${\text{K}}_c$.  The relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_c$ is given below.

${\text{K}}_{\text{p}}={\text{K}}_{\text{c}}{\text{RT}}^{\text{Δn}}$

Where,

R is the universal gas constant

T is the temperature

$\text{Δn}$ is the difference between the number of gaseous products to reactants.

Le Chatelier's  principle:  When a system at equilibrium is disturbed, the system will adjust itself to nullify the effect and maintain the equilibrium.

Answer to Problem 14.101QP

The equilibrium constant ${\text{K}}_{\text{p}}\text{=14}\text{.1}$

Explanation of Solution

Given,

The equilibrium constant ${\text{K}}_{\text{c}}=0.153$

$\begin{array}{l}\text{Temperature, T=}298\text{°C}\\ =850+273\\ =1123\text{K}\end{array}$

$\text{R}\text{=0}\text{.0821}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$

To find the difference between the number of gaseous products and reactants

${\text{C}}_{\text{(s)}}+{\text{CO}}_{\text{2(g)}}\rightleftharpoons {\text{2CO}}_{\text{(g)}}$

$\begin{array}{l}\text{Δn}=\text{Number}\text{of}\text{gaseous}\text{products}\text{-}\text{Number}\text{of}\text{gaseous}\text{reactants}\\ \text{= 2-1}\\ \text{=1}\end{array}$

To find ${\text{K}}_{\text{p}}$

The value of ${\text{K}}_{\text{p}}$ can be obtained by plugging in the values of ${\text{K}}_c$, $\text{Δn}$, $\text{R}$ and temperature in the given equation.

$\begin{array}{l}{\text{K}}_{\text{p}}={\text{K}}_{\text{c}}{\text{RT}}^{\text{Δn}}\\ =\left(0.153\right){\left(0.0821\times 1123\right)}^1\\ =14.1\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium constant has to be found and the equilibrium partial pressures of the reactant and product also has to be found.  The effect of decrease in temperature in the equilibrium constant has to be found.

Concept introduction:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where the reactant A is giving product B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Equilibrium constant $\left({\text{K}}_{\text{p}}\right)$: For gaseous substances the concentration will be proportional to its partial pressure.  The equilibrium constant for gaseous  reactions can be expressed in terms of the partial pressures of the gaseous products and reactants, and it is called equilibrium constant ${\text{K}}_{\text{p}}$  .  The value of ${\text{K}}_{\text{p}}$ is different from ${\text{K}}_c$.  The relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_c$ is given below.

${\text{K}}_{\text{p}}={\text{K}}_{\text{c}}{\text{RT}}^{\text{Δn}}$

Where,

R is the universal gas constant

T is the temperature

$\text{Δn}$ is the difference between the number of gaseous products to reactants.

Le Chatelier's  principle:  When a system at equilibrium is disturbed, the system will adjust itself to nullify the effect and maintain the equilibrium.

Answer to Problem 14.101QP

The partial pressure of ${\text{CO}}_2$ at equilibrium is found to be $\text{0}\text{.4atm}$

The partial pressure of $\text{CO}$ at equilibrium is found to be $\text{2}\text{.3atm}$

Explanation of Solution

Given,

The partial pressure of ${\text{CO}}_{\text{2}}=1.50\text{atm}$

The equilibrium constant ${\text{K}}_{\text{p}}\text{=14}\text{.1}$

To find pressure of reactant and product at equilibrium.

Using the table approach, the equilibrium concentrations of the reactants and the products can be found.

$\begin{array}{l}\text{Amount}\text{(M)}{\text{C}}_{\text{(s)}}+{\text{CO}}_{\text{2(g)}}\rightleftharpoons 2{\text{CO}}_{\text{(g)}}\\ \end{array}$

$\begin{array}{l}\text{Initial}1.50\rightleftharpoons \text{0}\text{}\\ \text{Change}\text{-x}\rightleftharpoons +2x\\ \text{Equillibrium}\left(1.50-x\right)\rightleftharpoons 2x\end{array}$

The equilibrium pressure values are then substituted into the equilibrium expression to get the change in pressure x.

${\text{K}}_p\text{=}\frac{{\text{P}}^{\text{2}}{}_{\text{CO}}}{{\text{P}}_{{\text{CO}}_{\text{2}}}}$

$14.10=\frac{{(2x)}^2}{(1.50-x)}$

On rearranging we get a quadratic equation.

$4x^2+14.10x-21.15=0$

On solving the quadratic equation the value of x obtained.

$x=\frac{-14.10\pm \sqrt{{(14.10)}^2-4(4)(-21.15)}}{2(4)}$

On solving we get two values for x, the positive value for x is taken.

$\text{x}=1.135$

Hence,

$\begin{array}{l}\text{The equilibrium pressure of}\text{CO}\text{=2x}\\ \text{=2}\times \text{1}\text{.135}\\ \text{= 2}\text{.3atm}\end{array}$

$\begin{array}{l}\text{The equilibrium pressure of}{\text{CO}}_2\text{=1}\text{.50-1}\text{.135}\\ \text{=0}\text{.4atm}\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium constant has to be found and the equilibrium partial pressures of the reactant and product also has to be found.  The effect of decrease in temperature in the equilibrium constant has to be found.

Concept introduction:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where the reactant A is giving product B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Equilibrium constant $\left({\text{K}}_{\text{p}}\right)$: For gaseous substances the concentration will be proportional to its partial pressure.  The equilibrium constant for gaseous  reactions can be expressed in terms of the partial pressures of the gaseous products and reactants, and it is called equilibrium constant ${\text{K}}_{\text{p}}$  .  The value of ${\text{K}}_{\text{p}}$ is different from ${\text{K}}_c$.  The relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_c$ is given below.

${\text{K}}_{\text{p}}={\text{K}}_{\text{c}}{\text{RT}}^{\text{Δn}}$

Where,

R is the universal gas constant

T is the temperature

$\text{Δn}$ Is the difference between the numbers of gaseous products to reactants.

Le Chatelier’s principle:  When a system at equilibrium is disturbed, the system will adjust itself to nullify the effect and maintain the equilibrium.

Answer to Problem 14.101QP

For and endothermic reaction the rate of forward reaction decreases with the decrease in temperature.  Hence, the reaction will shift towards the left and the partial pressure of carbon monoxide will be decreased

Explanation of Solution

According to Le Chatelier's principle a system at equilibrium is disturbed, it will try to get the equilibrium state back by nullifying the disturbance.  The given reaction is an endothermic reaction.  For and endothermic reaction the rate of forward reaction decreases with the decrease in temperature.  Hence, the reaction will shift towards the left and the partial pressure of carbon monoxide will be decreased.