\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$

Step by step solution

01

Identify the target reaction

The target reaction is \( C_{2}H_{2}(g) + 0.5O_{2}(g) \rightarrow 2CO_{2}(g) + H_{2}O(l) \). The purpose is to recognize that the enthalpy change for this reaction needs to be found.

02

Reorder the reactions

The reactions are studied and rearranged as follows: Reaction 1 remains unchanged \( C(s) + O_{2}(g) \rightarrow CO_{2}(g), \Delta H^{\circ} = -390 \, kJ/mol \) Reaction 2 is reversed to: \( H_{2}O(l) \rightarrow H_{2}(g) + 0.5O_{2}(g), \Delta H^{\circ} = +290 \, kJ/mol \) Reaction 3 is also reversed to: \( C_{2}H_{2}(g) \rightarrow 2C(s) + H_{2}(g), \Delta H^{\circ} = -230 \, kJ/mol \)

03

Combine the reactions

Systematically add up the revised reactions: \( C(s) + O_{2}(g) \rightarrow CO_{2}(g) \), \( H_{2}O(l) \rightarrow H_{2}(g) + 0.5O_{2}(g) \), and \( C_{2}H_{2}(g) \rightarrow 2C(s) + H_{2}(g) \). This yields to the desired reaction \( C_{2}H_{2}(g) + 0.5O_{2}(g) \rightarrow 2CO_{2}(g) + H_{2}O(l) \)

04

Calculate the final enthalpy change

The enthalpy change for the desired reaction, \(\Delta H_{target}\), is calculated by summing the enthalpy changes for each of the revised reactions. \[\Delta H_{target} = -390 kJ/mol + 290 kJ/mol - 230 kJ/mol = -330 kJ/mol \]

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