Chapter 12, Problem 96QRT

Interpretation Introduction

Interpretation:

From the equilibrium reaction of $\text{CO}\text{and}{\text{H}}_{\text{2}}\text{O}$ with ${\text{CO}}_{\text{2}}\text{and}{\text{H}}_{\text{2}}$, the cases at which the concentration of $\text{CO}$ increases and the concentration of $\text{CO}$ decreases has to be given.

Concept Introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ 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 A reacts to give B.

  $\text{aA}\rightleftharpoons \text{bB}$

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

On rearranging,

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

Where,

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

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

    ${\text{K}}_{\text{c}}$ is the equilibrium constant.

Reaction Quotient:

Reaction quotient, Q is the ratio between product of the product concentration to the product of the reactant concentration with each term raised to the power of its balancing coefficient.

Consider a general equation,

  $a\text{A +}b\text{A}\stackrel{}{\rightleftharpoons }c\text{C}+d\text{D}$

Where a, b, c and d are the stoichiometric coefficients.  The reaction quotient is,

  ${\text{Q}}_{\text{c}}\text{ = }\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{[\text{A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

The concentration of solids and pure liquids do not change, so their concentration terms are not included in the reaction quotient expression.

Comparison of $Q_{c}and{K}_c$:

When ${\text{Q}}_{\text{c}}{\text{ < K}}_{\text{c}}$, the reaction will proceed in right direction, $\text{reactants}\stackrel{}{\to }\text{products}$.

When ${\text{Q}}_{\text{c}}{\text{ > K}}_{\text{c}}$, the reaction will proceed in left direction, $\text{reactants}\stackrel{}{\leftarrow }\text{products}$.

When ${\text{Q}}_{\text{c}}{\text{ = K}}_{\text{c}}$, the reaction will remain unchanged, $\text{reactants}\stackrel{}{\rightleftharpoons }\text{products}$.

Explanation of Solution

Given information,

  Volume of flask is $\text{10}\text{.00}\text{L}$.

  Equilibrium constant value for the reaction,

  ${\text{CO}}_{\left(\text{g}\right)}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{CO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{H}}_{\text{2}\left(\text{g}\right)}{\text{K}}_{\text{C}}\text{=}\text{4}\text{.00}$

  $\begin{array}{l}\underset{\_}{n_{CO}\left(mol\right)n_{H_{2}O}\left(mol\right)n_{C{O}_2}\left(mol\right)n_{C{O}_2}\left(mol\right)}\\ Casea\text{1}\text{.0}\text{0}\text{.10}\text{0}\text{.10}\text{0}\text{.10}\\ Caseb\text{10}\text{1}\text{.0}\text{1}\text{.0}\text{1}\text{.0}\\ Casec\text{10}\text{10}\text{1}\text{.0}\text{1}\text{.0}\\ Cased\text{5}\text{.62}\text{0}\text{.381}\text{1}\text{.75}\text{1}\text{.75}\end{array}$

Case a:

Calculate the ${\text{Q}}_{\text{C}}$ value for the given reaction,

  ${\text{Q}}_{\text{C}}\text{=}\frac{\left[{\text{CO}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]}{\left[\text{CO}\right]\left[\text{HI}\right]}\text{=}\frac{\left(\frac{\text{0}\text{.10}\text{mol}{\text{CO}}_{\text{2}}}{\text{10}\text{.00}\text{L}}\right)\left(\frac{\text{0}\text{.10}\text{mol}{\text{H}}_{\text{2}}}{\text{10}\text{.00}\text{L}}\right)}{\left(\frac{\text{1}\text{.0}\text{mol}\text{CO}}{\text{10}\text{.00}\text{L}}\right)\left(\frac{\text{0}\text{.10}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{10}\text{.00}\text{L}}\right)}\text{=}\text{0}\text{.10}$

Compare ${\text{Q}}_{\text{C}}$ value with ${\text{K}}_{\text{C}}$ value,

  ${\text{Q}}_{\text{C}}\text{<}{\text{K}}_{\text{C}}$

Therefore, in case a, the concentration of $\text{CO}$ decreases.

Case b:

Calculate the ${\text{Q}}_{\text{C}}$ value for the given reaction,

  ${\text{Q}}_{\text{C}}\text{=}\frac{\left[{\text{CO}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]}{\left[\text{CO}\right]\left[\text{HI}\right]}\text{=}\frac{\left(\frac{\text{1}\text{.0}\text{mol}{\text{CO}}_{\text{2}}}{\text{10}\text{.00}\text{L}}\right)\left(\frac{1.0\text{mol}{\text{H}}_{\text{2}}}{\text{10}\text{.00}\text{L}}\right)}{\left(\frac{\text{10}\text{mol}\text{CO}}{\text{10}\text{.00}\text{L}}\right)\left(\frac{\text{1}\text{.0}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{10}\text{.00}\text{L}}\right)}\text{=}\text{0}\text{.10}$

Compare ${\text{Q}}_{\text{C}}$ value with ${\text{K}}_{\text{C}}$ value,

  ${\text{Q}}_{\text{C}}\text{<}{\text{K}}_{\text{C}}$

Therefore, in case b, the concentration of $\text{CO}$ decreases.

Case c:

Calculate the ${\text{Q}}_{\text{C}}$ value for the given reaction,

  ${\text{Q}}_{\text{C}}\text{=}\frac{\left[{\text{CO}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]}{\left[\text{CO}\right]\left[\text{HI}\right]}\text{=}\frac{\left(\frac{\text{1}\text{.0}\text{mol}{\text{CO}}_{\text{2}}}{\text{10}\text{.00}\text{L}}\right)\left(\frac{1.0\text{mol}{\text{H}}_{\text{2}}}{\text{10}\text{.00}\text{L}}\right)}{\left(\frac{\text{10}\text{mol}\text{CO}}{\text{10}\text{.00}\text{L}}\right)\left(\frac{\text{10}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{10}\text{.00}\text{L}}\right)}\text{=}\text{0}\text{.010}$

Compare ${\text{Q}}_{\text{C}}$ value with ${\text{K}}_{\text{C}}$ value,

  ${\text{Q}}_{\text{C}}\text{<}{\text{K}}_{\text{C}}$

Therefore, in case c, the concentration of $\text{CO}$ decreases.

Case d:

Calculate the ${\text{Q}}_{\text{C}}$ value for the given reaction,

  ${\text{Q}}_{\text{C}}\text{=}\frac{\left[{\text{CO}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]}{\left[\text{CO}\right]\left[\text{HI}\right]}\text{=}\frac{\left(\frac{\text{1}\text{.75}\text{mol}{\text{CO}}_{\text{2}}}{\text{10}\text{.00}\text{L}}\right)\left(\frac{\text{1}\text{.75}\text{mol}{\text{H}}_{\text{2}}}{\text{10}\text{.00}\text{L}}\right)}{\left(\frac{\text{5}\text{.62}\text{mol}\text{CO}}{\text{10}\text{.00}\text{L}}\right)\left(\frac{\text{0}\text{.381}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{10}\text{.00}\text{L}}\right)}\text{=}\text{1}\text{.43}$

Compare ${\text{Q}}_{\text{C}}$ value with ${\text{K}}_{\text{C}}$ value,

  ${\text{Q}}_{\text{C}}\text{<}{\text{K}}_{\text{C}}$

Therefore, in case d, the concentration of $\text{CO}$ decreases.