Chapter 14, Problem 14.27QA

Interpretation Introduction

To find:

The value of $K_c$ for the given reaction at temperature of the vessel from the given initial moles of reactant and moles of product at equilibrium.

Answer to Problem 14.27QA

Solution:

The value of $K_c$ for  $H_{2}O\left(g\right) + CO\left(g\right)\leftrightarrow H_2\left(g\right)+{CO}_2\left(g\right)$ from the given data is $1.5$.

Explanation of Solution

1) Concept:

The equilibrium constant expresses the relation between the amount of product and reactant at equilibrium.

The general expression for equilibrium constant $K_c$ is

$K_c= \frac{{\left[Product\right]}^p}{{\left[Reactant\right]}^r}$

where the exponents p and r are the stoichiometric coefficients of the product and reactant in the balanced equation.

ICE table can be used to calculate the concentration of the reactant and product at equilibrium from initial and equilibrium concentration.

2) Formula:

$K_c=\frac{\left[H_2\right]\left[{CO}_2\right]}{\left[H_{2}O\right]\left[CO\right]}$

3) Given:

i) $1.50\times {10}^{-2}mol of H_{2}O\left(g\right)$ (Initially)

ii) $1.50\times {10}^{-2}mol of CO\left(g\right)$(Initially)

iii) $8.3\times {10}^{-3}mol of {CO}_2\left(g\right)$(At equilibrium)

4) Calculation:

Assume the vessel volume to be1L.Then theinitial concentrations of the reactant and the equilibrium concentration of the product are:

$\left[H_{2}O\right]=\frac{1.50\times {10}^{-2} mol}{1 L}=1.50\times {10}^{-2}M$

$\left[CO\right]=\frac{1.50\times {10}^{-2} mol}{1 L}=1.50\times {10}^{-2}M$

$\left[{CO}_2\right]=\frac{8.3\times {10}^{-3} mol}{1 L}=8.3\times {10}^{-3}M$

Set up the ICE chart for the reaction:

$H_{2}O\left(g\right) + CO\left(g\right) \leftrightarrow H_2\left(g\right) + {CO}_2\left(g\right)$

Initial $1.50\times {10}^{-2} 1.50\times {10}^{-2} 0 0$

Change $-x -x +x +x$

Equilibrium $\left(1.50\times {10}^{-2}\right)-x \left(1.50\times {10}^{-2}\right)-xx x (=8.3\times {10}^{-3})$

Since H₂ and CO₂ have a 1:1 mole ratio, and their amounts at the start were zero, [H₂] = [CO₂] = 8.3×10^-3 M.

and $\left[H_{2}O\right]=\left[CO\right]=\left(1.50\times {10}^{-2}\right)-x=\left[\left(1.50\times {10}^{-2}\right)-\left(8.3\times {10}^{-3}\right)\right]=6.7\times {10}^{-3}M.$

Therefore  $K_c$ expression is

$K_c=\frac{\left[H_2\right]\left[{CO}_2\right]}{\left[H_{2}O\right]\left[CO\right]}=\frac{{\left(8.3\times {10}^{-3}\right)}^2}{{(6.7\times {10}^{-3})}^2}$

$K_c=\frac{6.889\times {10}^{-5}}{4.489\times {10}^{-5}}$

$K_c=1.535$

$K_c=1.5$ (in correct significant figures)

Conclusion:

We used ICE table for reactant’s initial concentration and concentration of product at equilibrium and calculated $K_c$ value from $K_c$ expression for given reaction.