From the following data, $$ \begin{array}{c} \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ calculate the enthalpy change for the reaction $$ 2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$

Step by step solution

01

Analyze the Target Reaction

First, we need to examine the reaction we are interested in, which is: \(2 C(graphite) + 3 H_{2}(g) \longrightarrow C_{2}H_{6}(g)\) In this reaction, we start with 2 moles of graphite and 3 moles of hydrogen gas, and end with 1 mole of ethane gas.

02

Identify Other Reactions

We are given three other reactions which involve these substances. We need these other reactions to calculate the enthalpy change of the target reaction, because enthalpy is a state function, so we can use the sum of the enthalpy changes of these reactions to find the enthalpy change of the target.

03

Rearrange Other Reactions

We need to rearrange these reactions so that we get our target reaction. From the third given reactions, we learn that ethane (C_{2}H_{6}) is formed from 4 CO_{2} and 6 H_{2}O. To match the target reaction, we have to revert this reaction and divide it by 2. Also, the second reaction needs to be divided by 2. So, the updated reactions and their enthalpies become: \[C(graphite) + O_{2}(g) \rightarrow CO_{2}(g), \(\Delta H_{rxn}^{\circ} = -393.5 ~kJ~/~mol\] \[H_{2}(g) + \frac{1}{2} O_{2}(g) \rightarrow H_{2}O(l), \(\Delta H_{rxn}^{\circ} = -285.8~/~2= -142.9 ~kJ ~ mol^{-1}\] \[C_{2}H_{6}(g) \rightarrow 2C(graphite) + 3 H_{2}(g), \(\Delta H_{rxn}^{\circ} = 3119.2 / 2 = 1559.8 ~kJ ~ mol^{-1}\]

04

Apply Hess' Law

Hess's Law says we can add up the enthalpy changes for these reactions to get the enthalpy change of the target reaction. In this case, the enthalpy change of the target reaction, \(\Delta H\), is \(\Delta H = \Delta H_{1} + \Delta H_{2} + \Delta H_{3} = -393.5 kJ~mol^{-1} - 142.9 kJ~mol^{-1} + 1559.8 kJ~mol^{-1} = 1023.4 kJ~mol^{-1}\)

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