Question bank/JEE Main/Chemistry

The equilibrium constant for the reaction,
SO3(g)[IMAGE] SO2(g) + $\frac{1}{2}$ O2(g)
is KC = 4.9 × 10-2. The value of KC for the reaction 2SO2(g) +
O2(g)[IMAGE] 2SO3(g) will be:
fuEu vfHkfØ;k] SO3(g)[IMAGE] SO2(g) + $\frac{1}{2}$ O2(g)
fLFkjkad KC = 4.9 × 10-2
O2(g)[IMAGE] 2SO3(g)

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SO3(g)[IMAGE] SO2(g) + O2(g)
$\frac{\left\lbrack {SO}_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack^{\frac{1}{2}}}{\left\lbrack {SO}_{3} \right\rbrack}$
=K_C = 4.9 × 10^-2......(i)
SO3(g) + 1/25 O2(g)[IMAGE] SO3(g)......(ii)
$\frac{\left\lbrack {SO}_{3} \right\rbrack}{\left\lbrack {SO}_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack^{1/2}} = K_{C}^{'} = \frac{1}{4.9 \times 10^{- 2}}$
For 2SO2(g) + O2(g)[IMAGE] 2SO3 (g)
$\frac{\left\lbrack {SO}_{3} \right\rbrack^{2}}{\left\lbrack {SO}_{2} \right\rbrack^{2}\left\lbrack O_{2} \right\rbrack} = K_{C}^{2} = \frac{1}{4.9 \times 4.9 \times 10^{- 4}} = \frac{10000}{24.01}$
= 416.49

Source: Grass-G / CH-27 Chemistry Paper 2 15 Jan.docx