The equilibrium constant for the reaction `H_(2)(g)+S(s) hArr H_(2)S(g)` is `18.5` at `925 K` and `9.25` at `1000 K`, respectively. Calculate the enthalpy of the reaction.

The equilibrium constant for the reaction `H_(2)(g)+S(s) hArr H_(2)S(g

1.	The equilibrium constant for the reaction `H_(2)(g)+S(s) hArr H_(2)S(g)` is `18.5` at `925 K` and `9.25` at `1000 K`, respectively. Calculate the enthalpy of the reaction.
Answer» Given `K_(1)=18.5` at `T_(1)=925` `K_(2)=9.25 at T_(2)=1`. Using the relation, `"log"K_(2)/K_(1)=(DeltaH)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `"log" 9.25/18.5=(DeltaH)/(2.303xx8.314)xx(75)/(925xx1000)` `-0.301=(DeltaHxx75)/(2.303xx8.314xx925xx1000)` or `DeltaH=-71080.57 J "mol"^(-1)`

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