Work done by the system in isothermal reversible process is `w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1))`. Also in case of adiabatic reversible process work done by the system is given by: `w_(rev.) = (nR)/(gamma -1) [T_2 - T_1]`. During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy `DeltaS = q_(rev.)/T`. The entropy changes also occurs during transformation of one state to other end expressed as `DeltaS = DeltaH/T`. Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and `DeltaG=DeltaH - T DeltaS`.
`Ag_2O_((s)) to 2Ag_((s)) + 1/2 O_(2(g))` attains equilibrium at temperature…`K` is : (The `DeltaH` and `DeltaS` for the reaction are `30.5kJ mol^(-1)` and `66J mol^(-1) K^(-1)` )

Options

A. `462.12`
B. 237
C. 373
D. 273

Answer

A

Explanation

At eq, `Delta G = 0 rArr Delta H - T Delta S = 0`
`T = (Delta H)/(Delta S)`