Question

For the reaction:
\[ \text{A(g)} \rightleftharpoons \text{B(g)} + \text{C(g)} \] Initial moles of A(g) is \(a\). At equilibrium, \(x\) moles of A decompose at total pressure \(P\). Calculate \(K_P\) for the given reaction.

• A. \(\dfrac{x^2P}{a^2-x^2}\)
• B. \(\dfrac{x^2P}{a^2+x^2}\)
• C. \(\dfrac{2xP}{a^2-x^2}\)
• D. \(\dfrac{xP}{a^2-x^2}\)

Hint:

For gaseous equilibrium problems, first write equilibrium moles, then convert them into mole fractions, and finally into partial pressures. After that, substitute directly into the \(K_P\) expression.

Answer 1

The Correct Option is A

Step 1: Write the equilibrium changes in moles.
The reaction is:
\[ \text{A(g)} \rightleftharpoons \text{B(g)} + \text{C(g)} \] Initially, moles of A \(= a\), and moles of B and C are zero.
If \(x\) moles of A decompose, then at equilibrium:
\[ \text{A} = a-x, \qquad \text{B} = x, \qquad \text{C} = x \] So, the total number of moles at equilibrium is:
\[ n_{\text{total}} = (a-x)+x+x = a+x \]
Step 2: Find the partial pressures of all gases.
If the total pressure at equilibrium is \(P\), then partial pressure of each gas is given by:
\[ p_i = \left(\frac{\text{moles of that gas}}{\text{total moles}}\right)P \] Therefore,
\[ P_A = \frac{a-x}{a+x}P \] \[ P_B = \frac{x}{a+x}P \] \[ P_C = \frac{x}{a+x}P \]
Step 3: Write the expression for \(K_P\).
For the reaction
\[ \text{A(g)} \rightleftharpoons \text{B(g)} + \text{C(g)} \] the equilibrium constant in terms of pressure is:
\[ K_P = \frac{P_B \cdot P_C}{P_A} \] Substituting the partial pressures:
\[ K_P = \frac{\left(\frac{x}{a+x}P\right)\left(\frac{x}{a+x}P\right)}{\left(\frac{a-x}{a+x}P\right)} \]
Step 4: Simplify the expression.
\[ K_P = \frac{x^2P^2}{(a+x)^2} \cdot \frac{a+x}{(a-x)P} \] Cancelling one \(P\) and one \((a+x)\):
\[ K_P = \frac{x^2P}{(a+x)(a-x)} \] Now use the identity:
\[ (a+x)(a-x)=a^2-x^2 \] Hence,
\[ K_P = \frac{x^2P}{a^2-x^2} \]
Step 5: Compare with the given options.
• (A) \(\dfrac{x^2P}{a^2-x^2}\): Correct. This matches the derived expression. • (B) \(\dfrac{x^2P}{a^2+x^2}\): Incorrect. The denominator should come from \((a+x)(a-x)\), which gives \(a^2-x^2\), not \(a^2+x^2\). • (C) \(\dfrac{2xP}{a^2-x^2}\): Incorrect. The numerator should contain \(x^2P\), not \(2xP\). • (D) \(\dfrac{xP}{a^2-x^2}\): Incorrect. One power of \(x\) is missing in the numerator.
Step 6: Conclusion.
Therefore, the correct value of \(K_P\) for the given reaction is:
\[ K_P=\frac{x^2P}{a^2-x^2} \] Final Answer: \(\dfrac{x^2P}{a^2-x^2}\)