JEE Advanced 2016 · Paper 2 · Chemistry Q34

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ChemistrySingle correct+3Question 34 of 54

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PARAGRAPH 1

Thermal decomposition of gaseous $\ce{X2}$ to gaseous X at 298 K takes place according to the following equation:
$\ce{X2(g) <=> 2X(g)}$
The standard reaction Gibbs energy, $\Delta_rG^\circ$ , of this reaction is positive. At the start of the reaction, there is one mole of $\ce{X2}$ and no X. As the reaction proceeds, the number of moles of X formed is given by $\beta$ . Thus, $\beta_{equilibrium}$ 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: $R=0.083\ \mathrm{L\,bar\,K^{-1}\,mol^{-1}}$ )

The INCORRECT statement among the following, for this reaction, is

Decrease in the total pressure will result in formation of more moles of gaseous X

At the start of the reaction, dissociation of gaseous

\ce{X2}

takes place spontaneously

\beta_{equilibrium}=0.7

K_C<1

Answer and worked solution

Official answer: C

(A) Lower pressure favours the side with more gas moles (2X). True.
(B) Initially $Q=0<K$ , so the forward reaction is spontaneous. True.
(C) $\Delta_rG^\circ>0$ means $K_p<1$ . But $\beta=0.7$ gives $K_p=\dfrac{8(0.49)}{4-0.49}=1.12>1$ . Incorrect.
(D) $K_C=K_p/(RT)$ with $K_p<1$ and $RT\approx24.7$ , so $K_C<1$ . True.

(C)

Topics:Equilibrium and Gibbs energy Le Chatelier's principle

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JEE Advanced 2016 · Paper 2, Chemistry question 34