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: Understanding the Concept:
K\(_P\) is the equilibrium constant expressed in terms of partial pressures of the gaseous reactants and products.
Partial pressure of a gas = (Mole fraction of that gas) \(\times\) (Total pressure).
For a reaction that proceeds with an increase in total number of moles, the mole fractions change at equilibrium, and this must be carefully accounted for when deriving K\(_P\).
Step 2: Key Formula or Approach:
1. Set up the equilibrium table (ICE table) to find the moles of each species at equilibrium.
2. Calculate the total moles at equilibrium.
3. Determine mole fractions and then partial pressures for each species.
4. Substitute into the K\(_P\) expression and simplify.
Step 3: Detailed Explanation:
Reaction: \(A(g) \rightleftharpoons B(g) + C(g)\)
Initial moles (t = 0):
Moles of A = \(a\), \quad Moles of B = 0, \quad Moles of C = 0
At equilibrium (\(t_{eq}\)):
Moles of A = \(a - x\)
Moles of B = \(x\)
Moles of C = \(x\)
Total moles at equilibrium (\(n_{total}\)): \[ n_{total} = (a - x) + x + x = a + x \]
Mole fractions at equilibrium:
\[ X_A = \frac{a - x}{a + x}, \quad X_B = \frac{x}{a + x}, \quad X_C = \frac{x}{a + x} \]
Partial pressures at total pressure P:
\[ P_A = \frac{a - x}{a + x} \cdot P, \quad P_B = \frac{x}{a + x} \cdot P, \quad P_C = \frac{x}{a + x} \cdot P \]
Expression for K\(_P\):
\[ K_P = \frac{P_B \cdot P_C}{P_A} = \frac{\left(\frac{x}{a + x} P\right)\left(\frac{x}{a + x} P\right)}{\left(\frac{a - x}{a + x} P\right)} \]
\[ K_P = \frac{x^2 \cdot P^2}{(a + x)^2} \cdot \frac{a + x}{(a - x) \cdot P} \]
\[ K_P = \frac{x^2 P}{(a + x)(a - x)} = \frac{x^2 P}{a^2 - x^2} \]
Step 4: Final Answer:
The value of K\(_P\) is \(\dfrac{x^{2}P}{a^{2} - x^{2}}\).