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The entropies of H(2)(g) and H(g) are 60...

The entropies of `H_(2)(g)` and `H(g)` are `60 " and " 50 J"mole"^(-1)K^(-1)` respectively at `300` K. Using the data given below calcualate the bond enthalpy of `H_(2)(g)` in Kcal `"mole"^(-1)`.
`" "H_(2)(g) rarr 2H(g)," "DeltaG^(@)=21.6 KJ "mole"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
8
`DeltaG^(@) = DeltaH^(@) - tDeltaS^(@)`
`21.6 = DeltaH_(H-H) - 300 (2 xx S_(H)^(@) - S_(H_(2))^(@))`
`DeltaH_(H-H) = 33.6 KJ "mole"^(-1) =8Kcal "mole"^(-1)`
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