find out the standard free energy change at `60^(@)C`and at 1 atn if the `N_(2)O_(4)` is 50 % dissociated

Options

A. `-800 .0 kJ mol^(-1)`
B. `+ 800.0 kJ mol^(-1)`
C. `789.89 JK^(-1) mol^(-1)`
D. `+ 789.98 JK^(-1)mol ^(-1)`

Answer

c

Explanation

`N_(2)O_(4)(g)Leftrightarrow2NO_(2)(g)`
` N_(2)O_(4)` is 50 % dissociated , the mole fraction of the substance is given by `x_(n2O4)= (1-05)/(1 + 0.5), x_(no2)= (2xx0.5)/(1+0.5)`
`P_(n2O4)=0.5/1.5xx 1 atm , p_(no2)1/1.5 xx 1atm`
the equilibrium constant ` K_(p)` is given by `k_(p) =(P_(no_(2)))^(2)/(P_(N_(2)O_(4)))=1.5/((1.5)^(2)(0.5))=1/0.75=1.33`
` Delta_(a)G^(Theta)=-RT in K_(p)`
`=- 8.314 Jk^(-1)mol^(-1)xx333 K xx 2.303xx 0.1239`
`=-789.89 JK^(-1) mol^(-1)`