Ionic EquilibriumHard

Question

The dissociation constant of NH3 at 27°C from the following data:

NH3 + H+ $\rightleftharpoons$­ NH4+; ∆H° = –52.21 kJ/mol; ΔS° = +1.6 JK–1 mol–1

H2O ­$\rightleftharpoons$ H+ + OH–; ∆H° = 54.70 kJ/mol; ΔS° = –76.3 JK–1 mol–1

Given: R = 8.3 J/K-mol

Options

A.e10
B.e−10
C.e−8
D.e−9

Solution

$NH_{3} + H_{2}O \rightleftharpoons NH_{4}^{+} + OH^{-}$

$\Delta H^{o} = ( - 52.21) + (54.70) = 2.49\text{ kJ} $$${\Delta S^{o} = 1.6 + ( - 76.3) = - 74.7\text{ J/K} }{\text{Now, }\Delta H^{o} = - RT.\ln K_{eq} }{\text{Or, 2490} - 300 \times ( - 74.7) = - 8.3 \times 300 \times \ln K_{eq} }{\therefore K_{eq} = e^{- 10}}$$

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