1.

Following reaction in equilibrium at 25^(@)C: 2NO(g,1 xx 10^(-5) "atm")+Cl_2(g,1 xx 10^(-2)"atm") harr 2NOCl(g, 1 xx 10^(-2) "atm") DeltaG^@ is

Answer»

`-45.65 KJ`
`-28.53 KJ`
`-22.82 KJ`
`-57.06 KJ`

Solution :`K_(P)=(P_(NOCL)^(2))/(P_(NO)^(2) xx P_(Cl_(2)))=((10^(-2))^(2))/((10^(-5))^(2) xx 10^(-2))=10^(8)`
`DeltaG^(@)= -RT` 1 N k = -45.65 kJ


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