Chapter 15, Problem 55GQ

The reaction of hydrogen and iodine to give hydrogen iodide has an equilibrium constant, K_c, of 56 at 435 °C.

1. (a) What is the value of K_p?
2. (b) Suppose you mix 0.045 mol of H₂ and 0.045 mol of I₂ in a 10.0-L flask at 425 °C. What is the total pressure of the mixture before and after equilibrium is achieved?
3. (c) What is the partial pressure of each gas at equilibrium?

Interpretation Introduction

Interpretation:

The value of ${\text{K}}_{\text{P}}$ in the reaction of hydrogen iodide formation has to be calculated.

Concept Introduction:

Equilibrium constant in terms of pressure${\text{[K}}_{\text{p}}\text{]}$: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

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

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}{\text{K}}_{\text{c}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{concentration}\\ {\text{K}}_{\text{p}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{pressure}\\ \text{Δn}\text{-}\text{change}\text{in}\text{number}\text{of}\text{moles}\end{array}$

Answer to Problem 55GQ

The value of ${\text{K}}_{\text{P}}$ in the given reaction is 56

Explanation of Solution

Given:

$\begin{array}{l}{\text{H}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{I}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{2HI}{}_{\text{(g)}}\\ {\text{K}}_{\text{C}}\text{=}\text{56}\\ \text{T}\text{=}\text{435°C}\text{=}\text{708}\text{K}\end{array}$

Using the equation ${\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}$, we can calculate the value of ${\text{K}}_{\text{P}}$.

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}\\ \text{=}\text{56×}{\left(\text{0}\text{.082×708}\right)}^{\text{0}}\text{since}\text{Δn}\text{=}\text{2-2}\text{=}\text{0}\\ \text{=}\text{56}\end{array}$

Interpretation Introduction

Interpretation:

The total pressure of the mixture before and after equilibrium has to be calculated.

Concept Introduction:

Equilibrium constant in terms of pressure${\text{[K}}_{\text{p}}\text{]}$: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

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

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}{\text{K}}_{\text{c}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{concentration}\\ {\text{K}}_{\text{p}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{pressure}\\ \text{Δn}\text{-}\text{change}\text{in}\text{number}\text{of}\text{moles}\end{array}$

Answer to Problem 55GQ

The total pressure of the mixture before equilibrium and after equilibrium is $\text{0}\text{.52}\text{atm}$.

Explanation of Solution

Given:

$\begin{array}{l}\text{No}\text{.}\text{of}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{=}\text{0}\text{.045}\text{mol}\\ \text{No}\text{.}\text{of}\text{moles}\text{of}{\text{I}}_{\text{2}}\text{=}\text{0}\text{.045}\text{mol}\\ \text{V}\text{=}\text{10}\text{L}\\ \text{T}\text{=}\text{425°C}\text{=}\text{698}\text{K}\\ \end{array}$

From the reaction ${\text{H}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{I}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{2HI}{}_{\text{(g)}}$, it is clear that one mole of hydrogen and one mole of iodine combine to form two ,moles of hydrogen iodide. Therefore

$\text{0}\text{.045}\text{mol}{\text{H}}_{\text{2}}\text{and}{\text{I}}_{\text{2}}\text{form}\text{0}\text{.09}\text{mol}\text{of}\text{HI}$

Total pressure can be calculated using ideal gas equation

$\text{PV}\text{=}\text{nRT}$

$\begin{array}{l}\text{P}\text{=}\frac{\text{n}\text{R}\text{T}}{\text{V}}\\ \text{=}\frac{\text{0}\text{.09}\text{×0}\text{.082×698}}{\text{10}}\\ \text{=}\text{0}\text{.52}\text{atm}\end{array}$

Total pressure of the mixture is $\text{0}\text{.52}\text{atm}$.

The total pressure before and after equilibrium will be the same since the amount of gas has not changed.

Interpretation Introduction

Interpretation:

The partial pressure of each gas in the mixture has to be calculated.

Concept Introduction:

Equilibrium constant in terms of pressure${\text{[K}}_{\text{p}}\text{]}$: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

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

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}{\text{K}}_{\text{c}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{concentration}\\ {\text{K}}_{\text{p}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{pressure}\\ \text{Δn}\text{-}\text{change}\text{in}\text{number}\text{of}\text{moles}\end{array}$

Answer to Problem 55GQ

The partial pressures of each gas in the mixture is $\begin{array}{l}{\text{P}}_{{\text{H}}_{\text{2}}}\text{=}{\text{P}}_{{\text{I}}_{\text{2}}}\text{=}\text{0}\text{.052}\\ {\text{P}}_{\text{HI}}\text{=}\text{0}\text{.12}\text{atm}\end{array}$

Explanation of Solution

Given that initially $\text{0}\text{.045}\text{mol}\text{of}{\text{H}}_{\text{2}}\text{and}{\text{I}}_{\text{2}}$ are present.

$\begin{array}{l}{\text{H}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{I}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{2HI}{}_{\text{(g)}}\\ \text{Initial}\text{concentration}\text{0}\text{.045}\text{mol}\text{0}\text{.045}\text{mol}\text{0}\\ \text{After}\text{reacting}\text{x}\text{moles}\text{0}\text{.045-x}\text{0}\text{.045-x}\text{2x}\end{array}$

For the reaction ${\text{H}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{I}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows \text{2HI}{}_{\text{(g)}}$

$\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\frac{{\left[\text{HI}\right]}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]}\\ \text{56}\text{=}\frac{{\left[\text{2x}\right]}^{\text{2}}}{{\left[\text{0}\text{.045-x}\right]}^{\text{2}}}\\ \text{56}\text{×}{\left[\text{0}\text{.045-x}\right]}^{\text{2}}\text{=}{\text{4x}}^{\text{2}}\\ {\text{52x}}^{\text{2}}\text{-5}\text{.04x}\text{+}\text{0}\text{.1134}\text{=}\text{0}\\ \text{x}\text{=}\text{0}\text{.06}\end{array}$

$\begin{array}{l}{\text{P}}_{{\text{H}}_{\text{2}}}\text{=}{\text{P}}_{{\text{I}}_{\text{2}}}\text{=}\text{0}\text{.052}\\ {\text{P}}_{\text{HI}}\text{=}\text{0}\text{.12}\text{atm}\end{array}$