ThermodynamicsHard
Question
A reaction at 300 K with ΔGo = −1743 J consists of 3 moles of A(g), 6 moles of B(g) and 3 moles of C(g). If A, B and C are in equilibrium in one liter vessel, then the reaction should be (ln 2 = 0.7, R = 8.3 J/K-mol)
Options
A.A + B $\rightleftharpoons$ C
B.A $\rightleftharpoons$ B + 2C
C.2A $\rightleftharpoons$ B + C
D.A + B $\rightleftharpoons$ 2C
Solution
$\Delta G^{o} = - RT.\ln K_{eq}$
$\Rightarrow - 1743 = - 8.3t300 \times \ln K_{eq} $$${\therefore K_{eq} = 2 }{\text{Now, }(a)K_{eq} = \frac{3}{3 \times 6}(b)K_{eq} = \frac{6 \times 3^{2}}{3} }{(c)K_{eq} = \frac{6 \times 3}{3^{2}}(d)K_{eq} = \frac{3^{2}}{3 \times 6}}$$
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