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In the following equilibrium N(2)O(4)(g...

In the following equilibrium
`N_(2)O_(4)(g) rarr 2NO_(2)(g)`
When 5 moles of each is taken and the temperature is kept at 298 k, the total pressure was found to be 20 bar
Given : `DeltaG^(@)_(f)(N_(2)O_(4) = 100 KJ)`
`DeltaG^(@)_(f)(NO_(2)) = 50 KJ`
Find `DeltaG^(@)` of the reaction at 298 K

A

`-4.68 KJ`

B

`-6.04 KJ`

C

`-5.705 KJ`

D

`0.4 KJ`

Text Solution

Verified by Experts

The correct Answer is:
C
Reaction quotient = `p[NO_(2)]^(2)/p[N_(2)O_(4)] = 100/10 = 10`
`DeltaG^(@) = 2DeltaG^(@)_(f)(NO_(2))-DeltaG^(@)_(f)(N_(2)O_(4)) = 2 xx 50 - 100 = 0`
`DeltaG = DeltaG^(@) - 2.303RT log_(10)Q_(p) = 0 - 2.303 xx 8.314 xx 298 log_(10)10 = -5705.8 J = -5.705 KJ`
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