Chapter 17, Problem 17.84P

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of ${\text{N}}_{\text{2}}{\text{ and H}}_{\text{2}}$ has to be calculated and $\text{Kc}$ for the given reaction has to be found.

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

Given, to a $\text{1}\text{.00-L}$ container at ${\text{727}}^{\text{°}}\text{C, 1}{\text{.30 mol of N}}_{\text{2}}\text{ and 1}{\text{.65 mol of H}}_{\text{2}}$ are added.  At equilibrium, $\text{0}\text{.100 mol}$ of ${\text{NH}}_{\text{3}}$ is present.

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

Explanation of Solution

Given reaction is

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

Initial:

Concentration of ${\text{NH}}_{\text{3}}$ is zero.

Concentration of ${\text{N}}_{\text{2}}$ is $\text{1}\text{.30 mol}$.

Concentration of ${\text{H}}_{\text{2}}$ is $\text{1}\text{.65 mol}$.

Change:

Concentration of ${\text{NH}}_{\text{3}}$ is $\text{+2x}$.

Concentration of ${\text{N}}_{\text{2}}$ is $\text{-x}$.

Concentration of ${\text{H}}_{\text{2}}$ is $\text{-3x}$.

At equilibrium:

Concentration of ${\text{NH}}_{\text{3}}$ is

  $\begin{array}{l}\text{2x = 0}\text{.100 M}\\ \\ \text{x = 0}\text{.0500 mol}\end{array}$

Concentration of ${\text{N}}_{\text{2}}$ is $\text{1}\text{.30 - x}$.

Concentration of ${\text{H}}_{\text{2}}$ is $\text{1}\text{.65 - 3x}$.

  $\begin{array}{l}{{\text{[N}}_{\text{2}}}_{\text{]}}\text{= (1}\text{.30-0}\text{.0500)M =1}\text{.25 M}\\ \\ {{\text{[H}}_{\text{2}}}_{\text{]}}\text{= (1}\text{.65-3(0}\text{.0500))M =1}\text{.50 M}\end{array}$

${\text{K}}_{\text{c}}$ can be calculated as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[N}}_{\text{2}}{\text{][H}}_{\text{2}}{\text{]}}^{\text{3}}}{{{\text{[NH}}_{\text{3}}\text{]}}^{\text{2}}}\\ \\ {\text{K}}_{\text{c}}\text{ = }\frac{\text{[1}\text{.25][1}{\text{.50]}}^{\text{3}}}{{\text{[0}\text{.100]}}^{\text{2}}}\\ \\ {\text{K}}_{\text{c}}\text{ = 421}\text{.875}\end{array}$

(b)

Interpretation Introduction

Interpretation:

For the given reaction, ${\text{K}}_{\text{c}}$ has to be calculated.

  ${\text{NH}}_{\text{3(g)}}\rightleftharpoons \frac{1}{2}{\text{N}}_{\text{2(g)}}\text{+ }\frac{\text{3}}{2}{\text{H}}_{\text{2(g)}}$

Given, in a different 1.00-L container at ${\text{727}}^{\text{°}}\text{C}$, equilibrium is established with $\text{8}{\text{.34×10}}^{\text{-2}}{\text{ mol of NH}}_{\text{3}}\text{, 1}{\text{.50 mol of N}}_{\text{2}}\text{ and 1}{\text{.25 mol of H}}_{\text{2}}$ present.

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

Explanation of Solution

${\text{K}}_{\text{c}}$ can to be calculated as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{{\text{[N}}_{\text{2}}\text{]}}^{\frac{\text{1}}{\text{2}}}{{\text{[H}}_{\text{2}}\text{]}}^{\frac{\text{3}}{\text{2}}}}{{\text{[NH}}_{\text{3}}\text{]}}\\ \\ {\text{K}}_{\text{c}}\text{ = }\frac{{\text{[1}\text{.50]}}^{\frac{\text{1}}{\text{2}}}{\text{[1}\text{.25]}}^{\frac{\text{3}}{\text{2}}}}{\text{[8}{\text{.34×10}}^{\text{-2}}\text{]}}\\ \\ {\text{K}}_{\text{c}}\text{ = 20}\text{.523177}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The relationship between the ${\text{K}}_{\text{c}}$ values in part (a) and part (b) has to be given and the reason why these values are not same has to be given.

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

Explanation of Solution

${\text{K}}_{\text{c}}$ in (a) is

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[N}}_{\text{2}}{\text{][H}}_{\text{2}}{\text{]}}^{\text{3}}}{{{\text{[NH}}_{\text{3}}\text{]}}^{\text{2}}}\\ \\ {\text{K}}_{\text{c}}\text{ = }\frac{\text{[1}\text{.25][1}{\text{.50]}}^{\text{3}}}{{\text{[0}\text{.100]}}^{\text{2}}}\\ \\ {\text{K}}_{\text{c}}\text{ = 421}\text{.875}\end{array}$

${\text{K}}_{\text{c}}$ in (b) is

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{{\text{[N}}_{\text{2}}\text{]}}^{\frac{\text{1}}{\text{2}}}{{\text{[H}}_{\text{2}}\text{]}}^{\frac{\text{3}}{\text{2}}}}{{\text{[NH}}_{\text{3}}\text{]}}\\ \\ {\text{K}}_{\text{c}}\text{ = }\frac{{\text{[1}\text{.50]}}^{\frac{\text{1}}{\text{2}}}{\text{[1}\text{.25]}}^{\frac{\text{3}}{\text{2}}}}{\text{[8}{\text{.34×10}}^{\text{-2}}\text{]}}\\ \\ {\text{K}}_{\text{c}}\text{ = 20}\text{.523177}\end{array}$

${\text{K}}_{\text{c}}$ in (a) is square of ${\text{K}}_{\text{c}}$ in (b).

It is because, the balanced equations of (a) and (b) are different.

(a)

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

(b)

  ${\text{NH}}_{\text{3(g)}}\rightleftharpoons \frac{1}{2}{\text{N}}_{\text{2(g)}}\text{+ }\frac{\text{3}}{2}{\text{H}}_{\text{2(g)}}$