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 Concept:
The equilibrium constants expressed in terms of partial pressure ($K_{p}$) and concentration ($K_{c}$) are related through the ideal gas law.
Step 2: Key Formula or Approach:
Use the relationship formula: $K_{p} = K_{c}(RT)^{\Delta n_{g}}$, rearranging it to solve for $K_{c}$: $K_{c} = K_{p}(RT)^{-\Delta n_{g}}$.
Calculate $\Delta n_{g} = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$, and convert temperature to Kelvin.
Step 3: Detailed Explanation:
1. Reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.

2. Calculate $\Delta n_{g$:}
$\Delta n_{g} = 2 \text{ (products)} - (1 + 3) \text{ (reactants)} = 2 - 4 = -2$.

3. Convert Temperature:
$T = 500^{\circ}\text{C} + 273 = 773$ K.

4. Gas Constant:
Since pressure is in atmosphere, use $R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}$.

5. Apply Formula:
\[ K_{p} = K_{c}(RT)^{\Delta n_{g}} \]
\[ K_{c} = K_{p}(RT)^{-\Delta n_{g}} \]
Substitute the values:
\[ K_{c} = 1.44 \times 10^{-5} \times (0.082 \times 773)^{-(-2)} \]
\[ K_{c} = 1.44 \times 10^{-5} \times (0.082 \times 773)^{2} \]
To match the specific format of the options, note that mathematically, $x^{2} = \frac{1}{x^{-2}}$. Therefore:
\[ K_{c} = \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}} \]
Step 4: Final Answer:
Matching this result with the given choices, (D) is the correct expression.