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ChemistrySingle correct+3Question 33 of 54

JEE Advanced 2016 · Paper 2, Chemistry question 33

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PARAGRAPH 1

Thermal decomposition of gaseous XX2\ce{X2} to gaseous X at 298 K takes place according to the following equation:
XX2(g)⇌2 X(g)\ce{X2(g) <=> 2X(g)}
The standard reaction Gibbs energy, ΔrG∘\Delta_rG^\circ, of this reaction is positive. At the start of the reaction, there is one mole of XX2\ce{X2} and no X. As the reaction proceeds, the number of moles of X formed is given by β\beta. Thus, βequilibrium\beta_{equilibrium} is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R=0.083 L bar K−1 mol−1R=0.083\ \mathrm{L\,bar\,K^{-1}\,mol^{-1}})

The equilibrium constant KPK_P for this reaction at 298 K, in terms of βequilibrium\beta_{equilibrium}, is

Choose one option
Answer and worked solution

Official answer: B

Moles at equilibrium. XX2\ce{X2}: 1−β21-\frac\beta2; X: β\beta; total 1+β21+\frac\beta2.

Partial pressures at 2 bar.
pX=2β1+β/2=4β2+β,pX2=2(2−β)2+β.p_X=\frac{2\beta}{1+\beta/2}=\frac{4\beta}{2+\beta},\qquad p_{X_2}=\frac{2(2-\beta)}{2+\beta}.

Equilibrium constant.
Kp=pX2pX2=16β2(2+β)2⋅2+β2(2−β)=8β24−β2.K_p=\frac{p_X^2}{p_{X_2}}=\frac{16\beta^2}{(2+\beta)^2}\cdot\frac{2+\beta}{2(2-\beta)}=\frac{8\beta^2}{4-\beta^2}.

(B)

Topics:Law of mass action and Kc, Kp
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