Question
Determine the standard enthalpy of the reaction C3H8(g) + H2(g) → C2H6(g) + CH4(g), using the given enthalpies under standard conditions.
Compound H2(g) CH4(g) C2H6(g) C (Graphite)
ΔcH0 (kJ/mol) −285.8 −890.0 −1560.0 −393.5
Options
Solution
$C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(l)$
$\Delta_{C}H_{C_{3}H_{3}(g)} = \left\lbrack 3 \times \Delta_{f}H_{CO_{2}(g)} + 4 \times \Delta_{f}H_{H_{2}O(l)} \right\rbrack - \left\lbrack \Delta_{f}H_{C_{3}H_{3}(g)} + 5 \times \Delta_{f}H_{O_{2}(g)} \right\rbrack $$$ = \left\lbrack 3( - 393.5) + 4( - 285.8) \right\rbrack - ( - 103.8) = - 2219.9\text{ kJ}$$
$\Delta_{r}H_{\text{required}} = - \left\lbrack \Delta_{C}H_{C_{2}H_{6}(g)} + \Delta_{C}H_{CH_{4}(g)} \right\rbrack + \left\lbrack \Delta_{C}H_{C_{3}H_{8}(g)} + \Delta_{C}H_{H_{2}(g)} \right\rbrack $$$ = \left\lbrack ( - 1560.0) + ( - 890.0) \right\rbrack + \left\lbrack ( - 2219.9) + ( - 285.8) \right\rbrack = - 55.7\text{ kJ}$$
Create a free account to view solution
View Solution Free