InterviewSolution
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InterviewSolution
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Calculate the enthalpy of formation of `Delta_(f)H` for `C_(2)H_(5)OH` from tabulated data and its heat of combustion as represented by the following equaitons: i. `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g), DeltaH^(Theta) =- 241.8 kJ mol^(-1)` ii. `C(s) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 393.5kJ mol^(-1)` iii. `C_(2)H_(5)OH (l) +3O_(2)(g) rarr 3H_(2)O(g) + 2CO_(2)(g), DeltaH^(Theta) =- 1234.7kJ mol^(-1)` a. `-2747.1 kJ mol^(-1)` b. `-277.7 kJ mol^(-1)` c. `277.7 kJ mol^(-1)` d. `2747.1 kJ mol^(-1)` |
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Answer» Required equation: `2C(s) +3H_(2)(g) +(1)/(2)O_(2)(g) rarr C_(2)H_(5)OH(l)` `2C(s) +2O_(2)(g) rarr 2CO_(2)(g), DeltaH^(Theta) = - 2xx 393.5 kJ` `3H_(2)(g) +O_(2)(g)rarr 3H_(2)O(g), DeltaH^(Theta) =- 3 xx 241.8 kJ` `3H_(2)O(l) +2CO_(2)(g) rarr C_(2)H_(5)OH(l) +3O_(2)(g),DeltaH^(Theta) =+ 1234.7 kJ` `ulbar({:(,"On adding,",,),(,"2C(s)+3H_(2)(g)+(1)/(2)O_(2)(g)rarrC_(2)H_(5)OH(l),,,),(,,DeltaH^(Theta)=-277.7 kJ,):})` |
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