Question

Study the reactions.

C (s)+O2 (g)→CO2 (g) ΔH=−394 kJ
H2 (g)+12O2 (g)→H2O (l) ΔH=−286 kJ
3C (s)+4H2 (g)→C3H8 (g) ΔH=106 kJ
Target Reaction:

C3H8 (g)+5O2 (g)→3CO2 (g)+4H2O (l)ΔH= ?
What is the enthalpy change of the target reaction?

Answer 1

Answer:

$\displaystyle \Delta H = -2426\text{ kJ}$

Explanation:

To find the enthalpy change of the target reaction, we can use Hess's Law.

Reversing the third reaction yields:

$\displaystyle \text{C _3 H _8 (g)} \longrightarrow 3\text{C(s)} + 4\text{H _2 (g)}\;\;\;\;\;\Delta H = -106\text{ kJ}$

Multiplying the first reaction by three yields:

$\displaystyle 3\text{C(s)} + 3\text{O _2 (g)} \longrightarrow 3\text{CO _2 }(g)}\;\;\;\;\; \Delta H = -1.18\times 10^3\text{ kJ}$

Multiplying the second reaction by four yields:

$\displaystyle 4\text{H _2 (g)} + 2\text{O _2 (g)} \longrightarrow 4\text{H _2 O( \ell )} \;\;\;\;\; \Delta H = -1.14\times 10^3 \text{ kJ}$

Adding all equations yield:

$\displaystyle \text{C _3 H _8 (g)} + 5\text{O _2 (g)} \longrightarrow 3\text{CO _2 (g)} + 4\text{H _2 O( \ell )} \;\;\;\;\; \Delta H = -2426\text{ kJ}$

Hence, the enthalpy change of the target reaction is -2426 kJ.