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For the reaction: N(2)O(4)(g) hArr 2NO(...

For the reaction: `N_(2)O_(4)(g) hArr 2NO_(2)(g)`
`(i)" "`In a mixture of `5 mol NO_(2)` and `5` mol `N_(2)O_(4)` and pressure of 20 bar. Calculate the value of `DeltaG` for the reaction. Given `DeltaG_(f)^(@) (NO_(2)) = 50` KJ/mol, `DeltaG_(f)^(@) ((N_(2)O_(4)) =100` KJ/mol and T=298 K. `(ii)` Predict the direction in which the reaction will shift, in order to attain equilibrium
[Given at `T=298 K, 2.303 "RT" = 5.7`KJ/mol.]

Text Solution

Verified by Experts

We need to calculate `Delta_(r )G`. Use : `Delta_(r ) G^(Θ)+` RT ln Q
So, first calculate `Delta_(r )G^(Θ)` using : `Delta_(r )G^(Θ)=sum G_("Products")^(Θ)-sum G_("Reactants")^(Θ)=sum (Delta_(f)G^(Θ))_("Products") - sum(Delta_(f)G^(Θ))_("Reactants")`
`[because "At standard conditions" : G_("compound")^(Θ)=Delta_(f)G_("compound")^(Θ)]`
`=2xxDelta_(f)G^(Θ)(NO_(2),g)-Delta_(f)G^(Θ)(N_(2)O_(4),g)=2xx50-100=0`
and `Delta_(r )G=Delta_(r )G^(Θ)+ RT ln Q = 0 8.314xx298 ln ((10^(2))/(10))=5.70 kJ mol^(-1) [because Q = (P_(NO_(2))^(2))/(P_(N_(2)O_(4)))=(((5)/(10)xx20)^(2))/(((5)/(10)xx20))]`
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