Question

At \(500\,\text{K}\), for the reaction, \[ N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g), \] the \(K_p\) is \(0.036\,\text{atm}^{-2}\). What is its \(K_c\) in \(L^2\text{mol}^{-2}\)? \((R=0.082\,L\,\text{atm mol}^{-1}\text{K}^{-1})\)

• A. \(2.1\times10^{-4}\)
• B. \(2.1\times10^{-5}\)
• C. \(60.5\)
• D. \(605\)

Hint:

For gaseous equilibrium, \[ K_p=K_c(RT)^{\Delta n} \] Always calculate \(\Delta n\) carefully as gaseous products minus gaseous reactants.

Answer 1

The Correct Option is C

Step 1: Write the relationship between $K_p$ and $K_c$.
For a gaseous reaction: \[ K_p = K_c (RT)^{\Delta n} \] where $\Delta n$ = moles of gaseous products minus moles of gaseous reactants.
Step 2: Calculate $\Delta n$ for the reaction.
For $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$: \[ \Delta n = 2 - (1 + 3) = -2 \]
Step 3: Rearrange to find $K_c$.
Since $K_p = K_c (RT)^{-2}$: \[ K_c = K_p \times (RT)^2 \]
Step 4: Calculate $RT$ and $(RT)^2$.
$R = 0.082$ L atm/mol/K, $T = 500$ K. \[ RT = 0.082 \times 500 = 41\ \text{L atm mol}^{-1} \] \[ (RT)^2 = 41^2 = 1681\ \text{L}^2\ \text{atm}^2\ \text{mol}^{-2} \]
Step 5: Substitute and calculate $K_c$.
$K_p = 0.036$ atm$^{-2}$. \[ K_c = 0.036 \times 1681 = 60.5\ \text{L}^2\ \text{mol}^{-2} \]
Step 6: State the final answer.
\[ \boxed{K_c = 60.5\ \text{L}^2\ \text{mol}^{-2}} \]