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

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

Step-by-step 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* 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* 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.