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

To find the equilibrium constant \(K_p\) for the reaction \(A(g) \rightleftharpoons B(g) + C(g)\), we will analyze the given situation and apply the concept of equilibrium and partial pressures.

Given:

• Initial amount of \(A(g)\) = \(a\)
• Equilibrium amount of \(A(g)\) = \(a - x\)
• Total pressure at equilibrium = \(p\)

To proceed, we will use the ideal gas law and the method of partial pressures. At equilibrium, the partial pressures of \(B(g)\) and \(C(g)\) are given by the equilibrium change.

Let's consider the changes in moles at equilibrium:

• Initial moles of \(A(g)\): \(a\)
• Change in moles of \(A(g)\): \(-x\) (since some \(A\) converts to \(B\) and \(C\))
• Moles of \(B(g)\) formed: \(+x\)
• Moles of \(C(g)\) formed: \(+x\)

Now, the total moles at equilibrium = \((a - x) + x + x = a + x\).

The partial pressures can be expressed as:

• Partial pressure of \(A\): \(\dfrac{a - x}{a + x} \times p\)
• Partial pressure of \(B\): \(\dfrac{x}{a + x} \times p\)
• Partial pressure of \(C\): \(\dfrac{x}{a + x} \times p\)

The equilibrium constant \(K_p\) is given by:

\(K_p = \dfrac{\text{(Partial Pressure of B)} \cdot \text{(Partial Pressure of C)}}{\text{(Partial Pressure of A)}}\)

Substitute the values:

\(K_p = \dfrac{\left(\dfrac{x}{a + x} \times p\right) \cdot \left(\dfrac{x}{a + x} \times p\right)}{\dfrac{a - x}{a + x} \times p}\)

\(K_p = \dfrac{x^2 \cdot p^2}{(a - x) \cdot p \cdot (a + x)}\)

Simplifying further, since the total ((a + x) is at the denominator and cancels out with the numerator partially):

\(K_p = \dfrac{x^2 \cdot p}{a^2 - x^2}\)

This matches with option:

\(\dfrac{x^2}{a^2 - x^2} \times p\)

 

Thus, the correct answer is \(\dfrac{x^2}{a^2 - x^2} \times p\).