`27^@C` पर अभिक्रिया,
`H_2(g) + 2Ag^(+)(aq) `Delta_r G` की गणना कीजिये।
दिया गया है :` P_(H_2) = 0.5. "bar" , [Ag^(+)] = 10^(-5) M`
`[H^+] = 10^(-3) M, Delta_f G^(@)[Ag^(+)(aq)]= 77.1 kJ//mol`

Options

A. `- 154.2 `kJ/mol
B. `- 178.9 `kJ/mol
C. `- 129.5 `kJ/मोल
D. इनमें से कोई नहीं

Answer

C

Explanation

`Delta_rG^(@) = 0 - 77.1 xx 2 = - 154.2 kJ//mol`
`Q = ([H^+]^2)/(P_(H_2)c[Ag^+]^2)`
`= (10^(-6))/(0.5 xx (10^(-10))) = 2 xx 10^4`
`DeltaG = Delta_rG^@ + RT ln Q`
`Delta_rG = -154.2 + (8.314 xx 300 ln(2 xx 10^4))/(1000)`
`= -129.5 kJ//mol`.