We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)$$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$\begin{array}{ll}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-890.3 \mathrm{kJ}} \\ {\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} & {\Delta H^{\circ}=-136.3 \mathrm{kJ}} \\ {2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-571.6 \mathrm{kJ}} \\ {2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-3120.8 \mathrm{kJ}}\end{array}$$

Step by step solution

01

Analyze the given reactions

We have the following reactions and their enthalpy changes: 1. \(CH_4(g) + 2 O_2(g) \longrightarrow CO_2(g) + 2 H_2O(l)\), \(\Delta H_1^\circ = -890.3 kJ\) 2. \(C_2H_4(g) + H_2(g) \longrightarrow C_2H_6(g)\), \(\Delta H_2^\circ = -136.3 kJ\) 3. \(2 H_2(g) + O_2(g) \longrightarrow 2 H_2O(l)\), \(\Delta H_3^\circ = -571.6 kJ\) 4. \(2 C_2H_6(g) + 7 O_2(g) \longrightarrow 4 CO_2(g) + 6 H_2O(l)\), \(\Delta H_4^\circ = -3120.8 kJ\)

02

Manipulate reactions to form the target reaction

We want to form the reaction: \[2 CH_4(g) \longrightarrow C_2H_4(g) + H_2(g) \] Manipulate the given reactions as follows: - Multiply reaction 1 by 2: \[2 CH_4(g) + 4 O_2(g) \longrightarrow 2 CO_2(g) + 4 H_2O(l), \Delta H_{1'}^\circ = -1780.6 kJ\] - Reverse reaction 2: \[C_2H_6(g) \longrightarrow C_2H_4(g) + H_2(g), \Delta H_{2'}^\circ = 136.3 kJ\] - Multiply reaction 3 by 1/2 and reverse it: \[H_2O(l) \longrightarrow H_2(g) + 1/2 O_2(g), \Delta H_{3'}^\circ = 285.8 kJ\] - Use reaction 4 as it is: \[2 C_2H_6(g) + 7 O_2(g) \longrightarrow 4 CO_2(g) + 6 H_2O(l), \Delta H_4^\circ = -3120.8 kJ\]

03

Add the manipulated reactions to obtain the target reaction

Now, add the manipulated reactions together: 1. \(2 CH_4(g) + 4 O_2(g) \longrightarrow 2 CO_2(g) + 4 H_2O(l)\) 2. \(C_2H_6(g) \longrightarrow C_2H_4(g) + H_2(g) \) 3. \(H_2O(l) \longrightarrow H_2(g) + 1/2 O_2(g) \) 4. \(2 C_2H_6(g) + 7 O_2(g) \longrightarrow 4 CO_2(g) + 6 H_2O(l)\) Sum: \(2 CH_4(g) + 4 CO_2(g) + 4 H_2O(l) \longrightarrow C_2H_4(g) + H_2(g) + 2 CO_2(g) + 6 H_2O(l) + 4 CO_2(g) \) Simplify: \(2 CH_4(g) \longrightarrow C_2H_4(g) + H_2(g)\)

04

Calculate the enthalpy change for the target reaction

Now, sum the \(\Delta H^\circ\) values of the manipulated reactions: \[\Delta H = \Delta H_{1'}^\circ + \Delta H_{2'}^\circ + \Delta H_{3'}^\circ + \Delta H_{4}^\circ\] \[\Delta H = (-1780.6 kJ) + (136.3 kJ) + (285.8 kJ) + (-3120.8 kJ)\] \[\Delta H = -3479.3 kJ\] So, the enthalpy change for the reaction \(2 CH_4(g) \longrightarrow C_2H_4(g) + H_2(g)\) is \(\Delta H^\circ = -3479.3 kJ\).

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