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The equilibrium constant for the reactio...

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.

Text Solution

Verified by Experts

According to avialable data:
`K_(1) = 18.5 T_(1) = 925 K, K_(2) = 1000 K, R = 8.314J`
`"Using the relation," log ((K_(2))/(K_(1)) )= (DeltaH)/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]=(DeltaH)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log((9.25)/(18.5)) = (DeltaH)/(2.303 xx 8.314)[(1000-925)/(1000xx925)]`
`-0.301 = (DeltaHxx75)/(2.303xx8.314xx1000xx925)`
`or " "DeltaH=(-0.301 xx 2. 303 xx 8. 314 xx 1000 xx 925)/(75) =- 71080.57 J`
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