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: Use the relation between \(K_p\) and \(K_c\).
For a gaseous reaction, \[ K_p=K_c(RT)^{\Delta n} \] where, \[ \Delta n=\text{moles of gaseous products}-\text{moles of gaseous reactants} \]

Step 2: Calculate \(\Delta n\).
For the reaction, \[ N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g) \] Moles of gaseous products: \[ 2 \] Moles of gaseous reactants: \[ 1+3=4 \] Therefore, \[ \Delta n=2-4 \] \[ \Delta n=-2 \]

Step 3: Substitute in the formula.
Since, \[ K_p=K_c(RT)^{-2} \] Therefore, \[ K_c=K_p(RT)^2 \] Given, \[ K_p=0.036 \] \[ R=0.082\,L\,\text{atm mol}^{-1}\text{K}^{-1} \] \[ T=500\,\text{K} \] Now, \[ RT=0.082\times500 \] \[ RT=41 \]

Step 4: Calculate \(K_c\).
\[ K_c=0.036\times(41)^2 \] \[ K_c=0.036\times1681 \] \[ K_c=60.516 \] \[ K_c\approx60.5 \]

Step 5: Final conclusion.
Hence, the value of \(K_c\) is \[ \boxed{60.5} \]