Question

The reaction $A(g) \rightleftharpoons B(g) + C(g)$ was initiated with the amount 'a' of $A(g)$. At equilibrium it is found that the amount of $A(g)$ remaining is $(a - x)$ at a total pressure of p.
The equilibrium constant $K_p$ of the reaction can be calculated from the expression :

• A. $\dfrac{x^2}{a^2 + x^2} \times p$
• B. $\dfrac{x^2}{a^2 - x^2} \times p$
• C. $\dfrac{a + x^2}{x^2} \times p$
• D. $\dfrac{a^2 - x^2}{x^2} \times p$

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
$K_p$ is defined as the product of the partial pressures of the products divided by the partial pressure of the reactant at equilibrium. Partial pressure of a gas is the product of its mole fraction and the total pressure.
: Key Formula or Approach:
\[ P_i = \chi_i \times P_{\text{total}} \]
\[ K_p = \frac{P_B \cdot P_C}{P_A} \]
Step 2: Detailed Explanation:
Let's set up the equilibrium table:
\begin{center} \begin{tabular}{lccc} & $A(g)$ & $\rightleftharpoons$ $B(g)$ & $+ \ C(g)$
Initial moles: & $a$ & 0 & 0
Change: & $-x$ & $+x$ & $+x$
At equilibrium: & $a-x$ & $x$ & $x$
\end{tabular} \end{center} Total moles at equilibrium $= (a - x) + x + x = a + x$.

Mole fractions at equilibrium:
$\chi_A = \frac{a - x}{a + x}$
$\chi_B = \frac{x}{a + x}$
$\chi_C = \frac{x}{a + x}$

Partial Pressures:
$P_A = \frac{a - x}{a + x} p$
$P_B = \frac{x}{a + x} p$
$P_C = \frac{x}{a + x} p$

Calculating $K_p$:
$K_p = \frac{\left( \frac{x}{a + x} p \right) \left( \frac{x}{a + x} p \right)}{\left( \frac{a - x}{a + x} p \right)} = \frac{x^2 p^2 / (a + x)^2}{(a - x)p / (a + x)}$
$K_p = \frac{x^2 p}{(a + x)(a - x)} = \frac{x^2}{a^2 - x^2} p$.
Step 3: Final Answer:
The expression for $K_p$ is $\frac{x^2}{a^2 - x^2} \times p$.