Question

Two moles each of the gases P$_2$, Q$_2$ and PQ are present in a vessel at equilibrium. If 1 mole each of P$_2$ and Q$_2$ is added at equilibrium, then determine the composition (in mole) of each species at the new equilibrium.

• A. $n_{P_2}=0.5,\ n_{Q_2}=0.5,\ n_{PQ}=1$
• B. $n_{P_2}=1.33,\ n_{Q_2}=1.33,\ n_{PQ}=1.67$
• C. $n_{P_2}=2.67,\ n_{Q_2}=2.67,\ n_{PQ}=2.33$
• D. $n_{P_2}=2.67,\ n_{Q_2}=2.67,\ n_{PQ}=2.67$

Hint:

When equilibrium constant is 1, products and reactants tend to equalize at equilibrium.

Answer 1

The Correct Option is D

Step 1: Write the equilibrium reaction.
\[ P_2 + Q_2 \rightleftharpoons 2PQ \]
Step 2: Initial equilibrium composition.
At equilibrium:
\[ n_{P_2} = 2,\quad n_{Q_2} = 2,\quad n_{PQ} = 2 \]
Step 3: Calculate equilibrium constant $K_c$.
\[ K_c = \frac{(n_{PQ})^2}{n_{P_2} \cdot n_{Q_2}} = \frac{(2)^2}{2 \times 2} = 1 \]
Step 4: After addition of reactants.
After adding 1 mole each of $P_2$ and $Q_2$:
\[ n_{P_2} = 3,\quad n_{Q_2} = 3,\quad n_{PQ} = 2 \]
Step 5: Let reaction proceed to new equilibrium.
Let $x$ moles of $P_2$ and $Q_2$ react:
\[ n_{P_2} = 3 - x,\quad n_{Q_2} = 3 - x,\quad n_{PQ} = 2 + 2x \]
Step 6: Apply equilibrium condition.
\[ K_c = \frac{(2+2x)^2}{(3-x)(3-x)} = 1 \]
Solving:
\[ 2 + 2x = 3 - x \Rightarrow x = \frac{1}{3} \]
Step 7: Final composition.
\[ n_{P_2} = 3 - \frac{1}{3} = 2.67 \] \[ n_{Q_2} = 3 - \frac{1}{3} = 2.67 \] \[ n_{PQ} = 2 + \frac{2}{3} = 2.67 \]