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For the equilibrium $\large 2{H_2}O\, \rightleftharpoons \,{H_3}{O^ + } + \,O{H^ - }$ the value of $\large \Delta {G^0}$ at 298 K is approximately :-

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-80 kJ mol^-1

-100 kJ mol^-1

100 kJ mol^-1

80 kJ mol^-1

answer is D.

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Detailed Solution

Variation of ΔG of a reaction is given as

ΔG = ΔG^o + RT ln Q

At equilibrium, ΔG = 0 and Q = K_i

ΔG^o = -RT ln K_i = -2.303 RT log K_i

ΔG^o = -2.303 x 8.3 x 298 log K_i

Since of the value of K_i is too small, K_w is used in place of K_i

ΔG^o = -2.303 x 8.3 x 298 x (-14)

ΔG^o = 79747.36 J

$\large \cong 80 kJ$

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