Calculate the heat of formation for propene $(C_3H_6)$ using the following thermochemical equations:
$(i) C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H_1 = -94.05 \ k.cal/mole$
$(ii) H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}; \Delta H_2 = -68.32 \ k.cal/mole$
$(iii) C_3H_{6(g)} + \frac{9}{2} O_{2(g)} \to 3 CO_{2(g)} + 3 H_2O_{(l)}; \Delta H_3 = -499.7 \ k.cal/mole$
(Note: The original question provided propane combustion data; assuming propene combustion data for consistency).
$(i) C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H_1 = -94.05 \ k.cal/mole$
$(ii) H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}; \Delta H_2 = -68.32 \ k.cal/mole$
$(iii) C_3H_{6(g)} + \frac{9}{2} O_{2(g)} \to 3 CO_{2(g)} + 3 H_2O_{(l)}; \Delta H_3 = -499.7 \ k.cal/mole$
(Note: The original question provided propane combustion data; assuming propene combustion data for consistency).
(A) The formation reaction for propene is: $3 C_{(s)} + 3 H_{2(g)} \to C_3H_{6(g)}$.
Using Hess's Law:
Multiply equation $(i)$ by $3$: $3 C_{(s)} + 3 O_{2(g)} \to 3 CO_{2(g)}$; $\Delta H = 3 \times (-94.05) = -282.15 \ k.cal/mole$.
Multiply equation $(ii)$ by $3$: $3 H_{2(g)} + \frac{3}{2} O_{2(g)} \to 3 H_2O_{(l)}$; $\Delta H = 3 \times (-68.32) = -204.96 \ k.cal/mole$.
Reverse equation $(iii)$: $3 CO_{2(g)} + 3 H_2O_{(l)} \to C_3H_{6(g)} + \frac{9}{2} O_{2(g)}$; $\Delta H = +499.7 \ k.cal/mole$.
Adding these equations:
$3 C_{(s)} + 3 H_{2(g)} + (3 + 1.5 - 4.5) O_{2(g)} \to C_3H_{6(g)}$
$\Delta H_f = -282.15 - 204.96 + 499.7 = +12.59 \ k.cal/mole$.
Using Hess's Law:
Multiply equation $(i)$ by $3$: $3 C_{(s)} + 3 O_{2(g)} \to 3 CO_{2(g)}$; $\Delta H = 3 \times (-94.05) = -282.15 \ k.cal/mole$.
Multiply equation $(ii)$ by $3$: $3 H_{2(g)} + \frac{3}{2} O_{2(g)} \to 3 H_2O_{(l)}$; $\Delta H = 3 \times (-68.32) = -204.96 \ k.cal/mole$.
Reverse equation $(iii)$: $3 CO_{2(g)} + 3 H_2O_{(l)} \to C_3H_{6(g)} + \frac{9}{2} O_{2(g)}$; $\Delta H = +499.7 \ k.cal/mole$.
Adding these equations:
$3 C_{(s)} + 3 H_{2(g)} + (3 + 1.5 - 4.5) O_{2(g)} \to C_3H_{6(g)}$
$\Delta H_f = -282.15 - 204.96 + 499.7 = +12.59 \ k.cal/mole$.
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