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)`A. `-2747.1kJ" "mol^(-1)`B. `-277.7kJ" "mol^(-1)`C. `277.7kJ" "mol^(-1)`D. `2747.1kJ" "mol^(-1)` |
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Answer» Correct Answer - B Required equation: `2C(s)+3H_(2)(g)+1//2O_(2)(g)toC_(2)H_(5)OH(l)` `2C(s)+2O_(2)(g)to2CO_(2)(g),DeltaH^(@)=-2xx393.5kJ` `3H_(2)(g)+(3)/(2)O_(2)(g)to3H_(2)O(g),DeltaH^(@)=3xx241.8kJ` `underline(3H_(2)O(l)+2CO_(2)(g)toC_(2)H_(5)OH(l)+3O_(2)(g)," "DeltaH=1234.7kJ)` On adding `underline(2C(s)+3H_(2)(g)+(1)/(2)O_(2)(g)toC_(2)H_(5)OH(l)," "DeltaH^(@)=-277.7kJ" "mol^(-1))` |
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