Question

For the reaction \( \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \), if equilibrium concentrations are:
\([ \text{N}_2 ] = 0.5\, \text{M}, \, [ \text{H}_2 ] = 1.5\, \text{M}, \, [ \text{NH}_3 ] = 1.0\, \text{M} \)
What is the value of \( K_c \)?

• A. 0.44
• B. 1.2
• C. 2.96
• D. 0.593

Hint:

Always raise each concentration to the power of its stoichiometric coefficient when writing equilibrium expressions.

Answer 1

The Correct Option is C

Step 1: Write the expression for equilibrium constant \( K_c \):
\[ K_c = \frac{[ \text{NH}_3 ]^2}{[ \text{N}_2 ][ \text{H}_2 ]^3} \] Step 2: Substitute the equilibrium concentrations:
\[ K_c = \frac{(1.0)^2}{(0.5)(1.5)^3} = \frac{1.0}{0.5 \times 3.375} = \frac{1.0}{1.6875} \approx 0.593 \] Step 3: Hence, the value of \( K_c \) is approximately 0.593.