Question

The following concentrations were obtained in the formation of $NH_{3(g)}$ from $N_{2(g)}$ and $H_{2(g)}$ at equilibrium at $500\ \text{K}$: $[NH_{3}]=1.5\times10^{-2}\ \text{M}$, $[N_{2}]=5\times10^{-3}\ \text{M}$ and $[H_{2}]=0.10\ \text{M}$. Calculate the equilibrium constant $K_{c}$ for the reaction $N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.

• A. 0.45
• B. 4.5
• C. 45.0
• D. $4.5\times10^{-2}$
• E. $4.5\times10^{-3}$

Hint:

Square the product concentration and cube the $H_{2}$ concentration per the coefficients.

Answer 1

The Correct Option is C

Step 1: Concept
$K_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$.

Step 2: Meaning
Plug the equilibrium concentrations into the expression.

Step 3: Analysis
$K_{c} = \frac{(1.5\times10^{-2})^{2}}{(5\times10^{-3})(0.10)^{3}} = \frac{2.25\times10^{-4}}{5\times10^{-3} \times 10^{-3}} = \frac{2.25\times10^{-4}}{5\times10^{-6}} = 45$.

Step 4: Conclusion
The equilibrium constant is 45.0. Final Answer: (C)