1.

Calculate enthalpy of formation of ethane at 25^(@)C if the enthalpies of combustion of carbon, hydrogen and ethane are 94.14, 68.47 and 373.3 kcal respectively.

Answer»

Solution :Given that,
(i) `{:(C(s)+O_(2)(g) to CO_(2)(g),,DeltaH=-94.14 "kcal"):}`
(ii) `{:(H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l),,DeltaH=-68.47"kcal"):}
(III) `{:(C_(2)H_(6)(g)+3(1)/(2)O_(2)(g) to 2CO_(2)(g)+3H_(2)O(l),,DeltaH=-373.3"kcal"):}`
We have to CALCULATE `DeltaH` of the equation,
(iv) `{:(2C(s)+3H_(2)(g) to C_(2)H_(6)(g),,DeltaH=?):}`
Carbon in Equation (i) and (iv), and hydrogen in equation (ii) and (iv) are on the same sides, but `C_(2)H_(6)` in equations (iii) and (iv) is on opposite sides. Thus following `[2xxEqn. (i) +3 xx Eqn. (ii) -Eqn. (iii)]`, we get,
`2C(s)+2O_(2)(g)+3H_(2)(g)+(3)/(2)O_(2)(g)-C_(2)H_(6)(g)-3(1)/(2)O_(2)(g) to 2CO_(2)(g)+3H_(2)O(l)-2CO_(2)(g)-3H_(2)O(l)-2CO_(2)(g)-3H_(2)O(l),`
`DeltaH=2xx(-94.14)+3xx(-68.47)-(-373.3)`
or `{:(2C(s)+3H_(2)(g) to C_(2)H_(6)(g),,DeltaH=-20.30"kcal"):}`


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