1.

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 barGiven : DeltaG^(@)_(f)(N_(2)O_(4) = 100 KJ)DeltaG^(@)_(f)(NO_(2)) = 50 KJFind DeltaG^(@) of the reaction at 298 K

Answer»

<P>`-4.68 KJ`
`-6.04 KJ`
`-5.705 KJ`
`0.4 KJ`

Solution :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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