Question

At equilibrium for the reaction N$_2$(g) + 3H$_2$(g) $\rightleftharpoons$ 2NH$_3$(g), the concentrations are [N$_2$] = 0.20 M, [H$_2$] = 0.60 M and [NH$_3$] = 0.80 M. The value of K$_c$ is:

• A. 7.4
• B. 14.8
• C. 22.2
• D. 44.4

Hint:

Always pay close attention to the stoichiometric coefficients in the chemical equation.
These coefficients become the exponents in the $K_c$ expression.
A common mistake is to forget to raise the concentration to its respective power (e.g., squaring $[\text{NH}_3]$ and cubing $[\text{H}_2]$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question is from the topic "Chemical Equilibrium," specifically involving the calculation of the equilibrium constant ($K_c$) for a given gaseous reaction using the given equilibrium concentrations of the reactants and products.

Step 2: Key Formula or Approach:
The equilibrium constant ($K_c$) is written as the ratio of the product of the equilibrium concentrations of the products to that of the reactants, with each concentration term raised to the power of its stoichiometric coefficient:
\[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]

Step 3: Detailed Explanation:

• We are given the equilibrium concentrations of the species involved in the reaction:
$[\text{N}_2] = 0.20\text{ M}$
$[\text{H}_2] = 0.60\text{ M}$
$[\text{NH}_3] = 0.80\text{ M}$

• Substituting these values into the equilibrium constant expression:
\[ K_c = \frac{(0.80)^2}{(0.20) \times (0.60)^3} \]

• Calculating the numerator:
\[ (0.80)^2 = 0.64 \]

• Calculating the denominator:
\[ (0.60)^3 = 0.216 \]
\[ 0.20 \times 0.216 = 0.0432 \]

• Substituting these calculated parts back into the expression:
\[ K_c = \frac{0.64}{0.0432} \]

• Simplifying the division:
\[ K_c \approx 14.81 \]



Step 4: Final Answer:
The value of the equilibrium constant $K_c$ is approximately $14.8$, which is option (B).