JEE Challenger
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ChemMedAdv 2016 • Paper 2

Equilibrium Constant Kp Expression in Moles of Products

Comprehension Passage

Thermal decomposition of gaseous X2X_2 into gaseous XX at 298 K298\text{ K} proceeds via the reaction: X2(g)⇌2X(g)X_2(g) \rightleftharpoons 2X(g) The standard reaction Gibbs free energy, ΔrG∘\Delta_r G^\circ, for this system is positive. Initially, 1 mole1\text{ mole} of X2X_2 is present without any XX. As decomposition progresses, the number of moles of XX produced is denoted by β\beta. Hence, βequilibrium\beta_{\text{equilibrium}} represents the moles of XX present at equilibrium. The process occurs under a constant total pressure of 2 bar2\text{ bar}. Assume ideal gas behavior. (Given: R=0.083 L bar K−1 mol−1R = 0.083\text{ L bar K}^{-1}\text{ mol}^{-1})

For this reaction at 298 K298\text{ K}, the expression for the equilibrium constant KpK_p in terms of βequilibrium\beta_{\text{equilibrium}} is given by

Options

A

8βequilibrium22−βequilibrium\frac{8\beta_{\text{equilibrium}}^2}{2 - \beta_{\text{equilibrium}}}

B

8βequilibrium24−βequilibrium2\frac{8\beta_{\text{equilibrium}}^2}{4 - \beta_{\text{equilibrium}}^2}

Correct
C

4βequilibrium22−βequilibrium\frac{4\beta_{\text{equilibrium}}^2}{2 - \beta_{\text{equilibrium}}}

D

4βequilibrium24−βequilibrium2\frac{4\beta_{\text{equilibrium}}^2}{4 - \beta_{\text{equilibrium}}^2}

Topics & Concepts

EquilibriumChemical Equilibrium

Step-by-Step Solution

To find the expression for the equilibrium constant KpK_p in terms of βequilibrium\beta_{\text{equilibrium}}, we analyze the given chemical equilibrium reaction:

X2(g)⇌2X(g)X_2(g) \rightleftharpoons 2X(g)

Step 1: Moles at Equilibrium

  • Initial moles of X2=1 molX_2 = 1\text{ mol}
  • Initial moles of X=0 molX = 0\text{ mol}

Let β=βequilibrium\beta = \beta_{\text{equilibrium}} be the number of moles of XX formed at equilibrium.

From the reaction stoichiometry, 1 mole of X2X_2 produces 2 moles of XX. Therefore, the number of moles of X2X_2 reacted to produce β\beta moles of XX is β2\frac{\beta}{2}.

At equilibrium:

  • Moles of X2X_2, nX2=1−β2=2−β2n_{X_2} = 1 - \frac{\beta}{2} = \frac{2 - \beta}{2}
  • Moles of XX, nX=βn_X = \beta

Step 2: Total Moles at Equilibrium The total number of moles of gas at equilibrium (ntotaln_{\text{total}}) is: ntotal=nX2+nX=(1−β2)+β=1+β2=2+β2n_{\text{total}} = n_{X_2} + n_X = \left(1 - \frac{\beta}{2}\right) + \beta = 1 + \frac{\beta}{2} = \frac{2 + \beta}{2}

Step 3: Partial Pressures at Equilibrium The total pressure given is P=2 barP = 2\text{ bar}.

The partial pressure of a gas is given by Pi=(nintotal)PP_i = \left(\frac{n_i}{n_{\text{total}}}\right) P.

  • Partial pressure of XX: PX=(β2+β2)×2=4β2+βP_X = \left( \frac{\beta}{\frac{2 + \beta}{2}} \right) \times 2 = \frac{4\beta}{2 + \beta}

  • Partial pressure of X2X_2: PX2=(2−β22+β2)×2=2(2−β)2+βP_{X_2} = \left( \frac{\frac{2 - \beta}{2}}{\frac{2 + \beta}{2}} \right) \times 2 = \frac{2(2 - \beta)}{2 + \beta}

Step 4: Calculation of Equilibrium Constant KpK_p The expression for KpK_p is: Kp=(PX)2PX2K_p = \frac{(P_X)^2}{P_{X_2}}

Substitute the values of partial pressures into the KpK_p expression: Kp=(4β2+β)22(2−β)2+βK_p = \frac{\left( \frac{4\beta}{2 + \beta} \right)^2}{\frac{2(2 - \beta)}{2 + \beta}}

Kp=16β2(2+β)22(2−β)2+βK_p = \frac{\frac{16\beta^2}{(2 + \beta)^2}}{\frac{2(2 - \beta)}{2 + \beta}}

Kp=16β2(2+β)2×2+β2(2−β)K_p = \frac{16\beta^2}{(2 + \beta)^2} \times \frac{2 + \beta}{2(2 - \beta)}

Kp=16β22(2+β)(2−β)K_p = \frac{16\beta^2}{2(2 + \beta)(2 - \beta)}

Using the algebraic identity (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2: Kp=8β24−β2K_p = \frac{8\beta^2}{4 - \beta^2}

Substituting β=βequilibrium\beta = \beta_{\text{equilibrium}} back into the expression: Kp=8βequilibrium24−βequilibrium2K_p = \frac{8\beta_{\text{equilibrium}}^2}{4 - \beta_{\text{equilibrium}}^2}

Hence, the correct option is B.

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