Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) given that $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+4 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ && \Delta H^{\circ}=-1937 \mathrm{kJ} \end{aligned}$$ $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.1 \mathrm{kJ} \end{array}$$

Step by step solution

01

Identify the target reaction.

The target reaction is: \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\).

02

Transform the first reaction to directly generate two \(H_2O\).

If we multiply the first reaction by 4, and then reverse it, it becomes: \(4 H_2O(l) \longrightarrow 2 H_2(g) + 2 O_2(g)\), \(\Delta H^{\circ}= + 1143.2 kJ\).

03

Transform the second reaction so it generates the right formula

We keep the second reaction same as it is: \(C_{3}H_{4}(g) + 4 O_{2}(g) \longrightarrow 3 CO_{2}(g) + 2 H_2O(l)\), \(\Delta H^{\circ}= -1937 kJ\).

04

Transform the third reaction so it gets rid of the unneeded formula

The third reaction is multiplied by -1: \(3 CO_{2}(g) + 4 H_2O(l) \longrightarrow C_{3}H_{8}(g) + 5 O_{2}(g)\), \(\Delta H^{\circ} = +2219.1 kJ\).

05

Add up the reactions

Add the reactions from Step 2, Step 3 and Step 4 together. Once added together, the \(O_2(g)\) molecules on the right side in Reactions 2 and 4 will cancel with those on the left side in Reaction 3. The \(H_2O(l)\) molecules on right side of Reactions 2 and 3 will cancel with \(H_2O(l)\) on the left side of reaction 4, the \(CO_{2}(g)\) molecules on right side of reaction3 will cancel with \(CO_{2}(g)\) on left side of reaction 4. And we end up with the reaction we want: \(C_{3}H_{4}(g) + 2 H_{2}(g) \longrightarrow C_{3}H_{8}(g)\). The \(\Delta H^{\circ}\) of this reaction is the sum of the \(\Delta H^{\circ}\) values from the modified reactions from Steps 2, 3, and 4: \(\Delta H^{\circ} = +1143.2 kJ – 1937 kJ + 2219.1 kJ = +1425.3 kJ\).

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