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For the formation of ammonia the equilib...

For the formation of ammonia the equilibrium constant data at `673K` and `773 K`, respectively, are `1.64xx10^(-4)` and `1.44xx10^(-5)` respectively. Calculate heat of reaction `(R=8.314 J K^(-1) mol^(-1))`

Text Solution

Verified by Experts

The given data are:
`K_(1)=1.64xx10^(-4), T_(1)=673 K`
`K_(2)=1.44xx10^(-5), T_(2)=773 K`
Using van't Hoff equation,
`log(K_(P_(2))/K_(P_(1)))=(DeltaH)/(2.303R)((T_(2)-T_(1))/(T_(2)T_(1)))`
`:. log (1.44xx10^(-5))/(1.64xx10^(-4))=(DeltaH)/(8.314xx2.303)((773-673)/(773xx673))`
`-1.0565=(DeltaH)/(19.147)(100/520229)`
`:. DeltaH=(-1.0565xx19.147xx520229)/(100)`
`=-105216 J= -105.216 kJ`
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