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

Identifying the Target Reaction

Identify the reaction for which the change in enthalpy is to be computed, which is: \(2C_{(graphite)} + 3H_2(g) \longrightarrow C_2H_6(g) \)

02

Rearranging the Given Reactions

Manipulate the given reactions in order to make them fit into the target reaction. First, reverse the third reaction and multiply it by 2: \(-2(C_2H_6(g) \longrightarrow 2C_{(graphite)} + 3H_2(g))\). This gives: \(2C_{(graphite)} + 3H_2(g) \longrightarrow 2C_2H_6(g)\). Second, reverse the first reaction: \(CO_2(g) \longrightarrow C_{(graphite)} + O_2(g)\). Third, keep the second reaction as is: \(H_2(g) + 0.5O_2(g) \longrightarrow H_2O(l)\)

03

Combining the Rearranged Reactions

Combining the manipulated reactions from step 2 gives: \(2C_{(graphite)}+3H_2(g) \longrightarrow 2C_2H_6(g)\), \(CO_2(g) \longrightarrow C_{(graphite)} + O_2(g)\), and \(6H_2(g) + 3O_2(g) \longrightarrow 6H_2O(l)\), which gives: \(2C_{(graphite)}+3H_2(g)+CO_2(g)+6H_2(g)+3O_2(g) \longrightarrow 2C_2H_6(g)+C_{(graphite)}+O_2(g)+6H_2O(l)\). Canceling out the same terms on both sides gives the final equation: \(2C_{(graphite)}+3H_2(g) \longrightarrow C_2H_6(g)\)

04

Calculating the Enthalpy Change

The change in enthalpy is computed by adding the enthalpy changes for each manipulated reaction. The enthalpy change of the first manipulated reaction is \(\Delta H_1 = -2 (-3119.6 kJ/mol) = 6239.2 kJ/mol\), of the second one is \(\Delta H_2 = -( - 393.5 kJ/mol) = 393.5 kJ/mol\), and of the third one is \(\Delta H_3 = 6(-285.8 kJ/mol) = -1714.8 kJ/mol\). The total change in enthalpy is \(\Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 = 6239.2 kJ/mol + 393.5 kJ/mol - 1714.8 kJ/mol = 4917.9 kJ/mol\)

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