Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements: 2C (graphite) + H2(g) → C2H2(g) The equations for each step and the corresponding enthalpy changes are

Question

Calculate the standard enthalpy of formation of acetylene ( [latex]C_{2}H_{2}[/latex]) from its elements:

2C (graphite) + H2(g) → [latex]C_{2}H_{2}[/latex](g)

The equations for each step and the corresponding enthalpy changes are

[latex]\begin{matrix} (a) & C(graphite)+O_{2}(g) ightarrow CO_{2}(g) & \Delta H ^{°}_{rxn}=-393.5 kJ/mol \ (b) & H_{2}(g)+\frac{1}{2}O_{2}(g) ightarrow H_{2}O(l) & \Delta H ^{°}_{rxn}=-285.8 kJ/mol \ (c) & 2C_{2}H_{2}(g)+5O_{2}(g) ightarrow 4CO_{2}(g)+2H_{2}O(l) & \Delta H ^{°}_{rxn}=-2598.8 kJ/mol \end{matrix} [/latex]

Strategy Our goal here is to calculate the enthalpy change for the formation of [latex]C_{2}H_{2}[/latex] from its elements C and [latex]H_{2}[/latex]. The reaction does not occur directly, however, so we must use an indirect route using the information given by Equations (a), (b), and (c).

Accepted Answer

Looking at the synthesis of [latex]C_{2}H_{2}[/latex] , we need 2 moles of graphite as reactant. So we multiply Equation (a) by 2 to get

[latex]\begin{matrix} (d) & 2C(graphite)+2O_{2}(g) ightarrow 2CO_{2}(g) & \Delta H^{°}_{rxn}=2(-393.5 kJ/mol) \ & & =-787.0 kJ/mol \end{matrix} [/latex]

Next, we need 1 mole of [latex]H_{2}[/latex] as a reactant and this is provided by Equation (b). Last, we need 1 mole of [latex]C_{2}H_{2}[/latex] as a product. Equation (c) has 2 moles of [latex]C_{2}H_{2}[/latex] as a reactant so we need to reverse the equation and divide it by 2:

[latex]\begin{matrix} (e) & 2CO_{2}(g)+H_{2}O(l) ightarrow C_{2}H_{2}(g)+\frac{5}{2}O_{2}(g) & \Delta H^{°}_{rxn}=\frac{1}{2} (2598.8 kJ/mol) \ & & =1299.4 kJ/mol \end{matrix}[/latex]

Adding Equations (d), (b), and (e) together, we get

[latex]\begin{matrix} 2C(graphite)+2O_{2}(g) ightarrow 2CO_{2}(g) & & \Delta H ^{°}_{rxn}=-787.0 kJ/mol \ H_{2}(g)+\frac{1}{2}O_{2}(g) ightarrow H_{2}O(l) & & \Delta H ^{°}_{rxn}=-285.8 kJ/mol \ 2CO_{2}(g)+2H_{2}O(l) ightarrow C_{2}H_{2}(g)+\frac{5}{2} O_{2}(g) & & \Delta H ^{°}_{rxn}=1299.4 kJ/mol \\hline 2CO_{2}(graphitie)+H_{2}(g) ightarrow C_{2}H_{2}(g) & & \Delta H ^{°}_{rxn}=226.6 kJ/mol \end{matrix} [/latex]

Therefore, [latex] \Delta H ^{°}_{f}=\Delta H ^{°}_{rxn}=226.6 kJ/mol[/latex] The [latex] \Delta H ^{°}_{f}[/latex] value means that when 1 mole of [latex]C_{2}H_{2}[/latex] is synthesized from 2 moles of C(graphite) and 1 mole of [latex]H_{2}[/latex] , 226.6 kJ of heat are absorbed by the reacting system from the surroundings. Thus, this is an endothermic process.

Question 6.9: Calculate the standard enthalpy of formation of acetylene (C...

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