JEE Main 2015 · 4 Apr · Shift 1 · Chemistry Q37

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The standard Gibbs energy change at 300 K for the reaction $\ce{2A <=> B + C}$ is 2494.2 J. At a given time, the composition of the reaction mixture is $[A]=\dfrac12$ , $[B]=2$ and $[C]=\dfrac12$ . The reaction proceeds in the : [R = 8.314 J/K/mol, e = 2.718]

forward direction because

Q>K_c

reverse direction because

Q>K_c

forward direction because

Q<K_c

reverse direction because

Q<K_c

Answer and worked solution

Official answer: B

$K_c=e^{-\Delta G^\circ/RT}=e^{-2494.2/(8.314\times300)}=e^{-1}\approx0.37$ .

$Q=\dfrac{[B][C]}{[A]^2}=\dfrac{2\times\frac12}{\frac14}=4$ . Since $Q>K_c$ , the reaction goes in the reverse direction.

Topics:Equilibrium and Gibbs energy Law of mass action and Kc, Kp

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JEE Main 2015 · 4 Apr · Shift 1, Chemistry question 37