Determine the standard enthlpy of the reaction C_(3)H_(8)(g) + H_(2)(g) to C_(2)H_(6)(g) + CH_(4)(g). Using the given enthalpies under standard conditions. Compound H_(2)(g)CH_(4)(g)C_(2)H_(6)(g)C("Graphite") Delta_(c)H^(0)(kJ//mol) –285.8 – 890.0 – 1560. 0 –395.5 The standard enthalpy of formation of C_(3)H_(8)(g) is –103.8 kJ//mol

Determine the standard enthlpy of the reaction C_(3)H_(8)(g) + H_(2)(g

1.	Determine the standard enthlpy of the reaction C_(3)H_(8)(g) + H_(2)(g) to C_(2)H_(6)(g) + CH_(4)(g). Using the given enthalpies under standard conditions. Compound H_(2)(g)CH_(4)(g)C_(2)H_(6)(g)C("Graphite") Delta_(c)H^(0)(kJ//mol) –285.8 – 890.0 – 1560. 0 –395.5 The standard enthalpy of formation of C_(3)H_(8)(g) is –103.8 kJ//mol
Answer» `-55.7 kJ` `+55.7 kJ` `-2060.4 kJ` `+ 2060.4 kJ` Solution :`C_(3)H_(8)(g) + 5O_(2)(g) to 3CO_(2)(g) + 4H_(2)O(l)` `Delta_(f)H_(H_(2)O(l)) = Delta_(C)H_(H_2(g))` `Delta_(C)H_(C_3H_3(g)) = [3 xx Delta H_(CO_2(g)) + 4 xx Delta_(f)H_(H_(2)O(l))]` `-[Delta_(f)H_(C_(3)H_(3)(g)) + 5 xx Delta_(f)H_(O_2(g))]` `=[3(-393.5)+4(-285.8)]` `-(-103.8) = -2219.9 kJ` `Delta_(f)H_("required") = -[Delta_(C)H_(C_2H_6(g)) + Delta_(C)H_(CH_4(g))]` `+[Delta_(C)H_(C_3H_8(g)) + Delta_(C)H_(H_2(g))]` `= -[(-1560.0) + (-890.0)]` `+[(-2219.9) + (-285.8)] = 55.7 kJ`.