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The standard enthalpy of formation of ga...

The standard enthalpy of formation of gaseous `H_(2)O` at 298 K is `-241.82` kJ/mol. Calculate `DeltaH^(@)` at 373 K given the following values of the molar heat capacities at constant pressure :
`H_(2)O(g)=33.58 " JK"^(-1)" mol"^(-1), " "H_(2)(g)=29.84 " JK"^(-1)" mol"^(-1), " "O_(2)(g)=29.37 " JK"^(-1)" mol"^(-1)`
Assume that the heat capacities are independent of temperature :

A

`-242.6 kJ/"mol"`

B

`+242.6 kJ//mol`

C

`+24.26` kJ/mol

D

`-242.6` J/mol

Text Solution

Verified by Experts

The correct Answer is:
A
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g), Delta_(r)H^(@) (298 K) = -241 kJ//mol^(-1)`
`Delta_(r)H_(373K)^(@) = Delta_(r)H_(298 K)^(@) + DeltaC_(p) (T_(2)-T_(1)) =-241.8 + [(C_(p))_(H_(2)O) -(C_(p))_(H_(2)) -1/2(C_(p))_(O_(2))] (373 - 298)`
`=-[241.8 + (33.6 - 28.8) - 1/2 xx 29.4] xx (373 - 298) xx 10^(-3) =-242.6 kJ//mol`
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