Chapter 7, Problem 7.104QP

(a)

Interpretation Introduction

Interpretation:

Equilibrium constant expression has to be written for the given reaction.

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

Concept Introduction:

Equilibrium constant is the one that relates the concentration of products and the reactants that is present in an equilibrium reaction.  In other words it can also be said that equilibrium constant is equal to the rate constant of forward reaction that is divided by the rate constant of reverse reaction.  Equilibrium constant is represented as $K_{eq}$.  General representation of equilibrium constant for a chemical reaction at equilibrium is,

    $\begin{array}{l}aA+bB\rightleftharpoons cC+dD\\ K_{eq}=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Equilibrium constant value has to be calculated for the given reaction if the concentration of ${\text{H}}_{\text{2}}\text{O}$ is $0.40M$, concentration of $\text{CO}$ is $0.40M$ and concentration of ${\text{H}}_{\text{2}}$ is $0.20M$.

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

Concept Introduction:

Refer part (a).