JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
The standard Gibbs energy change at \(300\, K\) for the reaction \(2A \rightleftharpoons B+C\) is \(2494.2\,J.\) At a given time, the composition of the reaction mixture is \([A] = \frac{1}{2}, [B] = 2\) and \([C] = \frac{1}{2} .\) The reaction proceeds in the : \([R = 8.314 \,J/K/mol, e= 2.718]\)
- A forward direction because \(Q < K_c\)
- B reverse direction because \(Q < K_c\)
- C forward direction because \(Q > K_c\)
- D reverse direction because \(Q > K_c\)
Answer & Solution
Correct Answer
(D) reverse direction because \(Q > K_c\)
Step-by-step Solution
Detailed explanation
\(\Delta G^{\circ}=2494.2 \,J\) \(2 A \rightleftharpoons B+C\) \(R=8.314 \,J / K / \mathrm{\,mol}\) \(e=2.718\) \([A]=\frac{1}{2},[B]=2\) \([C]=\frac{1}{2} ; Q=\frac{|B||C|}{|A|^{2}}\) \(=\frac{2 \times 1 / 2}{\left(\frac{1}{2}\right)^{2}}=4\)…
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