1.

find out the standard free energy change at60^(@)Cand at 1 atn if the N_(2)O_(4) is 50 % dissociated

Answer»

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

Solution :`N_(2)O_(4)(g)Leftrightarrow2NO_(2)(g)`
` N_(2)O_(4)` is 50 %dissociated , the mole fraction of thesubstance 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)`


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