Chapter 6, Problem 111CP

Interpretation Introduction

Interpretation: A plot for the variation of partial pressure of ${\text{NH}}_{\text{3}}\text{(g)}$ present at equilibrium varies with total pressures of $\text{1}\text{.0 atm, 10}\text{.0 atm, 100 atm and 1000 atm}$ needs to be drawn.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 111CP

  

Explanation of Solution

Given:

Initial moles for the ammonia synthesis reaction is: $\text{1}{\text{.0 mol N}}_{\text{2}}\text{(g) and 3}{\text{.0 mol H}}_{\text{2}}\text{(g)}$ .

At ${\text{450 }}^{\text{o}}{\text{C, K}}_{\text{p}}\text{ = 6}{\text{.5×10}}^{\text{-3}}$ .

The complete balanced reaction for the formation of ammonia is:

  ${\text{N}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}$

From the balanced reaction it is observed that 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas.

• For the total pressure of 1.0 atm:

Let the partial pressure of nitrogen gas be x atm so, the partial pressure of hydrogen gas will be 3x atm. Thus,

  $\text{x + 3x = 1}$ (sum of partial pressure is equal to the total pressure)

Solving for x:

  $\begin{array}{l}\text{4x = 1}\\ \text{x = }\frac{\text{1}}{\text{4}}\\ \text{x = 0}\text{.25 atm}\end{array}$

So, the partial pressure of nitrogen gas is $\text{0}\text{.25 atm}$ and the partial pressure of hydrogen gas will be $\text{3}\left(\text{0}\text{.25 atm}\right)\text{ = 0}\text{.75 atm}$ .

The ICE table for the reaction will be set up as:

  $\begin{array}{l}{\text{ N}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\\ \text{Initial(atm): 0}\text{.25 0}\text{.75 0}\\ \text{Change(atm): -x -3x +2x}\\ \text{Equilibrium(atm): 0}\text{.25-x 0}\text{.75-3x +2x}\end{array}$

The expression for the equilibrium constant is:

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\left({\text{p}}_{{\text{NH}}_{\text{3}}}\right)}^{\text{2}}}{\left({\text{p}}_{{\text{N}}_{\text{2}}}\right){\left({\text{p}}_{{\text{H}}_{\text{2}}}\right)}^{\text{3}}}$

Substituting the values:

  $\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{0}\text{.25-x}\right){\left(\text{0}\text{.75-3x}\right)}^{\text{3}}}$

Assuming the change be small so, $\text{0}\text{.25-x }\approx \text{ 0}\text{.25}$ and $\text{0}\text{.75-3x }\approx \text{ 0}\text{.75}$ thus,

  $\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{0}\text{.25}\right){\left(\text{0}\text{.75}\right)}^{\text{3}}}$

Solving for x:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\text{4x}}^{\text{2}}}{0.1055}\\ {\text{x}}^{\text{2}}\text{ = }\frac{\text{6}{\text{.5×10}}^{\text{-3}}\times 0.1055}{4}\\ \text{x = }\sqrt{\text{1}{\text{.7×10}}^{\text{-4}}}\\ \text{x = 1}{\text{.3×10}}^{\text{-2}}\end{array}$

The correct value of x is calculated by substituting the assumed x value as:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{0}\text{.25-0}\text{.013}\right){\left(\text{0}\text{.75-3}\left(\text{0}\text{.013}\right)\right)}^{\text{3}}}\\ \text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{\left({\text{4x}}^{\text{2}}\right)}{\left(\text{0}\text{.237}\right){\left(\text{0}\text{.711}\right)}^{\text{3}}}\\ {\text{x}}^{\text{2}}\text{ = }\frac{\text{6}{\text{.5×10}}^{\text{-3}}\text{× 0}\text{.0852}}{\text{4}}\\ \text{x = }\sqrt{\text{1}{\text{.38×10}}^{\text{-4}}}\\ \text{x = 1}{\text{.2×10}}^{\text{-2}}\end{array}$

The value of x is used and this process is repeated until the value of x does not vary:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{0}\text{.25-0}\text{.012}\right){\left(\text{0}\text{.75-3}\left(\text{0}\text{.012}\right)\right)}^{\text{3}}}\\ \text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{\left({\text{4x}}^{\text{2}}\right)}{\left(\text{0}\text{.238}\right){\left(\text{0}\text{.714}\right)}^{\text{3}}}\\ {\text{x}}^{\text{2}}\text{ = }\frac{\text{6}{\text{.5×10}}^{\text{-3}}\text{× 0}\text{.0866}}{\text{4}}\\ \text{x = }\sqrt{\text{1}{\text{.41×10}}^{\text{-4}}}\\ \text{x = 1}{\text{.2×10}}^{\text{-2}}\end{array}$

Since, the value x has not changed so, this value of x used. Thus, the partial pressure of ammonia is:

  ${\text{p}}_{{\text{NH}}_{\text{3}}}\text{ = 2}\left(\text{1}{\text{.2×10}}^{\text{-2}}\right)\text{ = 0}\text{.024 atm}$

• For the total pressure of 10.0 atm:

The ICE table for the reaction will be set up as:

  $\begin{array}{l}{\text{ N}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\\ \text{Initial(atm): 2}\text{.5 7}\text{.5 0}\\ \text{Change(atm): -x -3x +2x}\\ \text{Equilibrium(atm): 2}\text{.5-x 7}\text{.5-3x +2x}\end{array}$

The expression for the equilibrium constant is:

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\left({\text{p}}_{{\text{NH}}_{\text{3}}}\right)}^{\text{2}}}{\left({\text{p}}_{{\text{N}}_{\text{2}}}\right){\left({\text{p}}_{{\text{H}}_{\text{2}}}\right)}^{\text{3}}}$

Substituting the values:

  $\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{2}\text{.5-x}\right){\left(\text{7}\text{.5-3x}\right)}^{\text{3}}}$

Assuming the change be small so, $\text{2}\text{.5-x }\approx \text{ 2}\text{.5}$ and $\text{7}\text{.5-3x }\approx \text{ 7}\text{.5}$ thus,

  $\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{2}\text{.5}\right){\left(\text{7}\text{.5}\right)}^{\text{3}}}$

Solving for x:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\text{4x}}^{\text{2}}}{\text{1054}\text{.69}}\\ {\text{x}}^{\text{2}}\text{ = }\frac{\text{6}{\text{.5×10}}^{\text{-3}}\times \text{1054}\text{.69}}{4}\\ \text{x = }\sqrt{\text{1}\text{.71}}\\ \text{x = 1}\text{.3}\end{array}$

The correct value of x is calculated by substituting the assumed x value as:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{2}\text{.5-1}\text{.3}\right){\left(\text{7}\text{.5-3(1}\text{.3)}\right)}^{\text{3}}}\\ \text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{1}\text{.2}\right){\left(\text{3}\text{.6}\right)}^{\text{3}}}\\ {\text{x}}^{\text{2}}\text{ = }\frac{\text{6}{\text{.5×10}}^{\text{-3}}\times 55.99}{4}\\ \text{x = }\sqrt{\text{0}\text{.09}}\\ \text{x = 0}\text{.3}\end{array}$

The value of x is used and this process is repeated until the value of x does not vary. So, the calculated values of x are:

  $\begin{array}{l}\text{x = 1}\text{.0}\\ \text{x = 0}\text{.47}\\ \text{x = 0}\text{.86}\\ \text{x = 0}\text{.56}\\ \text{x = 0}\text{.79}\\ \text{x = 0}\text{.61}\\ \text{x = 0}\text{.75}\\ \text{x = 0}\text{.64}\\ \text{x = 0}\text{.72}\\ \text{x = 0}\text{.66}\\ \text{x = 0}\text{.71}\\ \text{x = 0}\text{.67}\\ \text{x = 0}\text{.70}\\ \text{x = 0}\text{.68}\\ \text{x = 0}\text{.69}\\ \text{x = 0}\text{.69}\end{array}$

Since, the value x has not changed so, this value of x used. Thus, the partial pressure of ammonia is:

  ${\text{p}}_{{\text{NH}}_{\text{3}}}\text{ = 2}\left(\text{0}\text{.69}\right)\text{ = 1}\text{.4 atm}$

• For the total pressure of 100.0 atm:

The ICE table for the reaction will be set up as:

  $\begin{array}{l}{\text{ N}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\\ \text{Initial(atm): 25 75 0}\\ \text{Change(atm): -x -3x +2x}\\ \text{Equilibrium(atm): 25-x 75-3x +2x}\end{array}$

The expression for the equilibrium constant is:

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\left({\text{p}}_{{\text{NH}}_{\text{3}}}\right)}^{\text{2}}}{\left({\text{p}}_{{\text{N}}_{\text{2}}}\right){\left({\text{p}}_{{\text{H}}_{\text{2}}}\right)}^{\text{3}}}$

Substituting the values:

  $\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{25-x}\right){\left(\text{75-3x}\right)}^{\text{3}}}$

The value of x is calculated using successive approximations as:

  $\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2}\right)}^{\text{2}}}{\left(\text{25}\right){\left(\text{75-3x}\right)}^{\text{3}}}$

Solving for x:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{\text{4}}{\left(\text{25}\right){\left(\text{75-3x}\right)}^{\text{3}}}\\ \text{x = 24}\end{array}$

The correct value of x is calculated by substituting the assumed x value as:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2}\times \text{24}\right)}^{\text{2}}}{\left(\text{25-24}\right){\left(\text{75-3x}\right)}^{\text{3}}}\\ \text{x = 1}\text{.4}\end{array}$

The value of x is used and this process is repeated until the value of x does not vary. So, the calculated values of x are:

  $\begin{array}{l}\text{x = 23}\text{.7}\\ \text{x = 3}\text{.6}\\ \text{x = 22}\text{.6}\\ \text{x = 8}\text{.1}\\ \text{x = 20}\text{.5}\\ \text{x = 12}\text{.1}\\ \text{x = 18}\text{.6}\\ \text{x = 14}\text{.7}\\ \text{x = 17}\text{.2}\\ \text{x = 15}\text{.5}\\ \text{x = 16}\text{.7}\\ \text{x = 15}\text{.8}\\ \text{x = 16}\text{.5}\end{array}$

Since, the value x has 16 is repeated so, this value of x used. Thus, the partial pressure of ammonia is:

  ${\text{p}}_{{\text{NH}}_{\text{3}}}\text{ = 2}\left(\text{16}\right)\text{ = 32 atm}$

• For the total pressure of 1000.0 atm:

The ICE table for the reaction will be set up as:

  $\begin{array}{l}{\text{ N}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\\ \text{Initial(atm): 250 750 0}\\ \text{Change(atm): -x -3x +2x}\\ \text{Equilibrium(atm): 250-x 750-3x +2x}\end{array}$

The expression for the equilibrium constant is:

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\left({\text{p}}_{{\text{NH}}_{\text{3}}}\right)}^{\text{2}}}{\left({\text{p}}_{{\text{N}}_{\text{2}}}\right){\left({\text{p}}_{{\text{H}}_{\text{2}}}\right)}^{\text{3}}}$

Substituting the values:

  $\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{250-x}\right){\left(\text{750-3x}\right)}^{\text{3}}}$

The value of x is calculated using successive approximations as:

  $\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2}\right)}^{\text{2}}}{\left(\text{250}\right){\left(\text{750-3x}\right)}^{\text{3}}}$

Solving for x:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{\text{4}}{\left(\text{250}\right){\left(\text{750-3x}\right)}^{\text{3}}}\\ \text{x = 249}\text{.5}\end{array}$

The correct value of x is calculated by substituting the assumed x value as:

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{{\left(\text{2}\times \text{249}\text{.5}\right)}^{\text{2}}}{\left(\text{25-249}\text{.5}\right){\left(\text{75-3x}\right)}^{\text{3}}}\\ \text{x = 110}\end{array}$

The value of x is used and this process is repeated until the value of x does not vary. So, the calculated values of x are:

  $\begin{array}{l}\text{x = 237}\\ \text{x = 204}\\ \text{x = 223}\\ \text{x = 815}\end{array}$

Since, according to significant digits, the value x has 220 is repeated so, this value of x used. Thus, the partial pressure of ammonia is:

  ${\text{p}}_{{\text{NH}}_{\text{3}}}\text{ = 2}\left(\text{220}\right)\text{ = 440 atm}$

Thus, the values of partial pressure of ammonia at total pressures are:

total pressure (atm)	partial pressure of ammonia(atm)
1.0	0.024
10.0	1.4
100.0	32
1000.0	440

Plotting the data as: