Question

For the reversible reaction, 
\( \text{N}_2 (g) + 3\text{H}_2 (g) \rightleftharpoons 2\text{NH}_3 (g) \). 
When the partial pressure is measured in atmosphere, the value of \( K_p \) at \( 500^\circ\text{C} \) is \( 1.44 \times 10^{-5} \). 
The value of \( K_c \) when the concentration is expressed in \( \text{mol L}^{-1} \) is:
\(\underline{\hspace{3cm}}\).

• A. \( \frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}} \)
• B. \( \frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}} \)
• C. \( \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{2}} \)
• D. \( \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}} \)

Hint:

Always double-check the sign of \( \Delta n_g \). For the Haber process (ammonia synthesis), \( \Delta n_g \) is always \(-2\). Also, remember that in \( K_p \)/\( K_c \) relations, temperature must always be in Kelvin.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to find the numerical expression for the equilibrium constant in terms of molarity (\( K_c \)) given the equilibrium constant in terms of partial pressure (\( K_p \)) and the temperature.
Step 2: Key Formula or Approach:
The relationship between \( K_p \) and \( K_c \) for a gaseous reaction is:
\[ K_p = K_c(RT)^{\Delta n_g} \] Where:
- \( R \) is the universal gas constant (use \( 0.0821 \, \text{L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \) to match units).
- \( T \) is the absolute temperature in Kelvin.
- \( \Delta n_g \) is the change in the number of moles of gaseous products and reactants (\( \text{moles of products} - \text{moles of reactants} \)).
Step 3: Detailed Explanation:
1. Calculate \(\Delta n_g\):
The balanced chemical equation is: \( \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \).
\[ \Delta n_g = (2) - (1 + 3) = 2 - 4 = -2 \]
2. Convert Temperature to Kelvin:
\[ T = 500^\circ\text{C} + 273 = 773 \, \text{K} \]
3. Rearrange the formula to solve for \( K_c \):
\[ K_c = \frac{K_p}{(RT)^{\Delta n_g}} \]
Substituting the values (\( K_p = 1.44 \times 10^{-5} \), \( R \approx 0.082 \), \( T = 773 \), and \( \Delta n_g = -2 \)):
\[ K_c = \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}} \]
Step 4: Final Answer:
Comparing this result with the given options, it exactly matches option (D). (Note: Option C is incorrect because it uses a positive exponent of 2 in the denominator, which would correspond to \( \Delta n_g = +2 \)).