Question

What is the numerical value of the equilibrium constant Kc?

Answer 1

ANSWER

EXPLANATION

Given that;

The number of moles of NH3 is 3 moles

The number of moles of H2 is 1 mole

The number of moles of N2 is 2 moles

At equilibrium, the concentration of ammonia is 1.4 moles/L

To find the value of Kc, follow the steps below

Step 1: Write the balanced equation of the reaction

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

Step 2: Write the equation of the reaction in terms of Kc

$\text{ K}_C\text{ = }\frac{[\text{ NH}_3]^2}{[N_2]\text{ \lbrack H}_2]^3}$

Step 3: Find the concentration of each reactant at equilibrium using a stoichiometry ratio

From the reaction above, you will see that 1 mole of Nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia.

Let x represents the concentration of nitrogen at equilibrium

Recall, that the concentration of ammonia at equilibrium is 1.4 moles/L

$\begin{gathered} \text{ 2 mole NH}_3\text{ }\rightarrow\text{ 1 mole N}_2 \\ \text{ 1.04 mole/L NH}_3\text{ }\rightarrow\text{ x moles/L N}_2 \\ \text{ cross multiply} \\ \text{ 2 moles NH}_3\times\text{ x moles/L N}_2\text{ = 1 mole N}_2\times1.4\text{ mol/L} \\ \text{ Isolate x} \\ \text{ x mol/L N}_2\text{ = }\frac{1\text{ moles N}_2\times1.41\cancel{\frac{mol}{L}}}{2\cancel{moles}} \\ \text{ x mol/L = }\frac{1.04}{2} \\ \text{ x= 0.52 mole/L} \end{gathered}$

Since the initial number of moles of nitrogen is 1 mole, hence, the concentration of nitrogen at equilibrium is calculated below as

Concentration at equilibrium = 1 -0.52

Concentration of nitrogen at equilibrium = 0.48 mole/L

The next step is to find the concentration of hydrogen at equilibrium

Let y represent the mole of hydrogen at equilibrium

$\begin{gathered} \text{ 2 moles NH}_3\rightarrow\text{ 3 moles H}_2 \\ \text{ 1.04 moles/L NH}_3\text{ }\rightarrow\text{ y moles/L H}_2 \\ \text{ cross multiply} \\ \text{ 2 moles NH}_3\times\text{ y moles/L H}_2\text{ = 1.04 moles/L NH}_3\times3\text{ moles H}_2 \\ \text{ Isolate y} \\ \text{ y moles/L H}_2\text{ = }\frac{1.04\cancel{\frac{moles}{L}}NH_3\times3mole\text{ H}_2}{2\cancel{molsNH_3}} \\ y\text{ = }\frac{1.40\times3}{2} \\ \text{ y = }\frac{3.12}{2} \\ \text{ y = 1.56 moles} \end{gathered}$

Since the initial concentration of hydrogen is 2 moles, hence the concentration of hydrogen at equilibrium can be calculated below as

Concentration at equilibrium = 2 - 1.56

Concentration at equilibrium = 0.44 mole/L

Step 4: Find the value of Kc using the equation in step 2

$\text{ Kc = }\frac{[NH_3]^2}{[N_2]\text{ \lbrack H}_2]^3}$$\begin{gathered} \text{ Kc = }\frac{1.04^2}{0.48\times0.44^3} \\ \\ \text{ K}_c=\text{ }\frac{1.0816}{0.48\times\text{ 0.085184}} \\ \\ \text{ kc = }\frac{1.0816}{0.0408832} \\ \text{ kc = 26.4558547276} \end{gathered}$

Hence, the value of Kc is 26