Q17.83CP - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the formation of ammonia in two experiments.

1.   To a $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{L}$ container at ${\mathbf{727}}^{\mathbf{o}}\mathbf{ }\mathbf{C}$ $\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{ }\mathbf{mol}$ of ${\mathbf{N}}_{\mathbf{2}}$ and $\mathbf{1}\mathbf{.}\mathbf{65}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}$ are added. At equilibrium, $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{NH}}_{\mathbf{3}}$  is present. Calculate the equilibrium concentrations of ${\mathbf{N}}_{\mathbf{2}}\mathbf{ }\mathbf{and}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}$, and find ${\mathbf{K}}_{\mathbf{c}}$ for the reaction: $\mathbf{2}{\mathbf{NH}}_{\mathbf{3}}\left(g\right)\mathbf{\to }{\mathbf{N}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\left(g\right)$  
2.   In a different $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{L}$ container at the same temperature, equilibrium is established with $\mathbf{8}\mathbf{.}\mathbf{34}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}\mathbf{ }}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{NH}}_{\mathbf{3}}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}\mathbf{,}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}$  present. Calculate ${\mathbf{K}}_{\mathbf{c}}$ for the reaction: ${\mathbf{NH}}_{\mathbf{3}}\left(g\right)\mathbf{\to }{\mathbf{N}}_{\mathbf{2}}\left(g\right)\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{H}}_{\mathbf{2}}\left(g\right)$ 
3.  (c) What is the relationship between the ${\mathbf{K}}_{\mathbf{c}}$  values in parts  $\left(a\right)$ and $\left(b\right)$ ? Why aren't these values the same?

Step-by-Step Solution

Answer

1. The concentrations of  ${\mathrm{N}}_2 \mathrm{and} {\mathrm{H}}_2$ at equilibrium are $1.25 \mathrm{M}$ and $1.50 \mathrm{M}$ respectively.
2.  The ${\mathrm{K}}_{\mathrm{c}}$ is $20.5$ .
3.   The relationship is given as ${\mathrm{K}}_{\mathrm{c}2}=\sqrt{{\mathrm{K}}_{\mathrm{c}1}} \mathrm{or} {\mathrm{K}}_{\mathrm{c}1}={\left({\mathrm{K}}_{\mathrm{c}2}\right)}^2$. The values are not same because the mole ratios of $\left(\mathrm{a}\right) \mathrm{and} \left(\mathrm{b}\right)$ varies.

Step 1: Concept Introduction

Commercial ammonia is made by catalysing the reaction of nitrogen and hydrogen at high temperatures and pressures. Fritz Haber and Carl Bosch, two German chemists, invented the technique in 1909.

Step 2: Calculating the molarity of [ N 2 ] , [ H 2 ] and [ NH 3 ]

(a) To calculate the equilibrium concentrations given reaction $2{\mathrm{NH}}_3\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)$ , we need to first identify the given values.

$$\begin{gathered} \mathrm{V}=1.00 \mathrm{L} \\ \mathrm{T}={727}^{\mathrm{o}} \mathrm{C} \\ =1000 \mathrm{K} \\ {{\mathrm{n}}_{\mathrm{N}}}_2=1.30 \mathrm{mol} \\ {{\mathrm{n}}_{\mathrm{H}}}_2=1.65 \mathrm{mol} \end{gathered}$$

${{\mathrm{n}}_{\mathrm{NH}}}_3=0.100 \mathrm{mol}$ (at equilibrium)

Solve for the molarity of each reaction component

 $\begin{array}{rcl}\left[{\mathrm{N}}_2\right]& =& \frac{1.30 \mathrm{mol}}{1.00 \mathrm{L}}\\ & =& 1.30 \mathrm{Mn}\\ \left[{\mathrm{H}}_2\right]& =& \frac{1.65 \mathrm{mol}}{1.00 \mathrm{L}}\\ & =& 1.65 \mathrm{Mn}\\ \left[{\mathrm{NH}}_3\right]& =& \frac{0.100 \mathrm{mol}}{1.00 \mathrm{L}}\\ & =& 0.100 \mathrm{Mn}\end{array}$

 

The molarity of $\left[{\mathrm{N}}_2\right]$, $\left[{\mathrm{H}}_2\right]$ and $\left[{\mathrm{NH}}_3\right]$ is $1.3 \mathrm{Mn}$, $1.65 \mathrm{Mn}$ and $0.100 \mathrm{Mn}$ respectively.

 

Now,  Calculating the concentration of N₂ and H₂ at equilibrium

Writing the reaction table

Now, solving for x ,

$$\begin{gathered} 0.100=0-2\mathrm{x} \\ \mathrm{x}=-\frac{0.100}{2} \\ \mathrm{x}=-0.0500 \end{gathered}$$ 

Now solve for the concentration of  ${\mathrm{N}}_2 \mathrm{and} {\mathrm{H}}_2$ at equilibrium.

 $\begin{array}{rcl}\left[{\mathrm{N}}_2\right]& =& 1.3+\mathrm{x}\\ & =& 1.3+\left(-0.050\right)\\ \left[{\mathrm{N}}_2\right]& =& 1.25 \mathrm{M}\\ \left[{\mathrm{H}}_2\right]& =& 1.65+3\mathrm{x}\\ & =& 1.65+3\left(-0.050\right)\\ \left[{\mathrm{H}}_2\right]& =& 1.50 \mathrm{M}\\ & & \end{array}$

The concentrations of  ${\mathrm{N}}_2 \mathrm{and} {\mathrm{H}}_2$ at equilibrium are $1.25 \mathrm{M}$ and $1.50 \mathrm{M}$ respectively.

 

Now, finding the value of ${\mathrm{K}}_{\mathrm{c}}$

To find the value of ${\mathrm{K}}_{\mathrm{c}}$ we need to write the expression for the equilibrium constant of the reaction.

 $$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{\left[\mathrm{product}\right]}{\left[\mathrm{reactant}\right]} \\ {\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}{{\left[{\mathrm{NH}}_3\right]}^2} \end{gathered}$$

Now, solving for the ${\mathrm{K}}_{\mathrm{c}}$ ,

 $$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}{{\left[{\mathrm{NH}}_3\right]}^2} \\ {\mathrm{K}}_{\mathrm{c}}=\frac{\left(1.25\right){\left(1.50\right)}^3}{{\left(0.100\right)}^2} \\ {\mathrm{K}}_{\mathrm{c}}=422 \end{gathered}$$

Therefore the value of ${\mathrm{K}}_{\mathrm{c}}$ is given as 422. 

Step 5: Calculating equilibrium concentration of NH 3 ( g ) → 1 2 N 2 ( g ) + 3 2 H 2 ( g )

(b)  To calculate the equilibrium concentration of the given reaction ${\mathrm{NH}}_3\left(\mathrm{g}\right)\to \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)$ , we first need to identify the given values.

$\begin{array}{rcl}\mathrm{V}& =& 1.00 \mathrm{L}\\ \mathrm{T}& =& {727}^{\mathrm{o}} \mathrm{C}\\ & =& 1000.15 \mathrm{K}\\ {{\mathrm{n}}_{\mathrm{NH}}}_3& =& 8.34\times {10}^{-2} \mathrm{mol} \left(\mathrm{at} \mathrm{equilibrium}\right)\\ {{\mathrm{n}}_{\mathrm{N}}}_2& =& 1.50 \mathrm{mol} \left(\mathrm{at} \mathrm{equilibrium}\right)\\ {{\mathrm{n}}_{\mathrm{H}}}_2& =& 1.25 \mathrm{mol} \left(\mathrm{at} \mathrm{equilibrium}\right)\end{array}$ 

Solve for the molarity of each reaction component.

$\begin{array}{rcl} \left[{\mathrm{NH}}_3\right]& =& \frac{8.34\times {10}^{-2} \mathrm{mol}}{1.00 \mathrm{L}}\\ & =& 8.34\times {10}^{-2} \mathrm{M} \left(\mathrm{at} \mathrm{equilibrium}\right)\\ \left[{\mathrm{H}}_2\right]& =& \frac{1.50 \mathrm{mol}}{1.00 \mathrm{L}}\\ & =& 1.50 \mathrm{M} \left(\mathrm{at} \mathrm{equilibrium}\right)\\ \left[{\mathrm{H}}_2\right]& =& \frac{1.25 \mathrm{mol}}{1.00 \mathrm{L}}\\ & =& 1.25 \mathrm{M} \left(\mathrm{at} \mathrm{equilibrium}\right)\end{array}$ 

 

 

Now, finding the value of ${\mathrm{K}}_{\mathrm{c}}$ 

To find the value of ${\mathrm{K}}_{\mathrm{c}}$ we need to write the expression for the equilibrium constant of the reaction.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]} \\ {\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\mathrm{N}}_2\right]}^{{\displaystyle \frac{1}{2}}}{\left[{\mathrm{H}}_2\right]}^{{\displaystyle \frac{3}{2}}}}{\left[{\mathrm{NH}}_3\right]} \end{gathered}$$ 

Solve for ${\mathrm{K}}_{\mathrm{c}}$ ,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\mathrm{N}}_2\right]}^{{\displaystyle \frac{1}{2}}}{\left[{\mathrm{H}}_2\right]}^{{\displaystyle \frac{3}{2}}}}{\left[{\mathrm{NH}}_3\right]} \\ {\mathrm{K}}_{\mathrm{c}}=\frac{{\left(1.50\right)}^{{\displaystyle \frac{1}{2}}}{\left(1.25\right)}^{{\displaystyle \frac{3}{2}}}}{\left(8.34\times {10}^{-2}\right)} \\ {\mathrm{K}}_{\mathrm{c}}=20.5 \end{gathered}$$ 

Therefore we get the value of ${\mathrm{K}}_{\mathrm{c}}$ as 20.5 .

Step 7: Finding the relationship between the K c values in parts ( a )   and   ( b ) ( a )   and   ( b )

(c)  Because the mole ratios of (a) and (b) varies, the result will be different and thus will naturally yield a different K_C value. Since they are the K_C of the same reaction, there is a relationship between them. When dealing with reactions that alter their mole ratios, the equilibrium constant can be easily modified. The K of the initial equation will be raised to n when the mole ratio increases by a number m. When a mole ratio falls by a number m , we take the ${\mathrm{m}}^{\mathrm{th}}$ root of the K of the original equation.

The mole ratio of equation (1) is 2 times the mole ratio of equation (2) as can be seen.

 $\left(1\right)=2\mathrm{eq}\left(2\right)$

Since it takes 2 eq 2  to have eq 1 , the ${\mathrm{K}}_{\mathrm{c}2}=\sqrt{{\mathrm{K}}_{\mathrm{c}1}} \mathrm{or} {\mathrm{K}}_{\mathrm{c}1}={\mathrm{K}}_{\mathrm{c}2}^2$.