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Chapter 9: Problem 84

Using this information: $$ \begin{array}{rr} \mathrm{C}(s) \longrightarrow \mathrm{C}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=716 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{H}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=872.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ and the fact that the average \(\mathrm{C}-\mathrm{H}\) bond enthalpy is \(414 \mathrm{~kJ} / \mathrm{mol}\), estimate the standard enthalpy of formation of methane \(\left(\mathrm{CH}_{4}\right)\).

Short Answer

Expert verified
The estimated standard enthalpy of formation of methane is \(-67.2 \, kJ/mol.\)

Step by step solution

01

Formulate the target reaction

The standard enthalpy of formation of CH4 involves converting elemental carbon and hydrogen into methane. This can be represented as follows: \n\( \mathrm{C}(s) + 2\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g) \)
02

Consider given reactions and enthalpies

We are given three pieces of enthalpy information: \n(i) The enthalpy change to convert solid carbon to gaseous carbon, \(\Delta H_{C(s) \to C(g)} = 716 \, kJ/mol.\)\n(ii) The enthalpy change to convert 2 moles of gaseous hydrogen to 4 moles of atomic hydrogen, \(\Delta H_{2H_{2}(g) \to 4H(g)} = 872.8 \, kJ/mol.\) Note that we ultimately need only 2 moles of atomic hydrogen, so we can halve this value.\n(iii) The bond enthalpy for a C-H bond, \(\Delta H_{C-H bond} = 414 \, kJ/mol.\) Since methane has 4 such bonds, we can multiply this value by 4.
03

Apply Hess's Law

Hess's Law allows us to add together chemical equations and their enthalpies to formulate new reactions and find their enthalpies. \nStart with \(\Delta H_{C(s) \to C(g)}.\) \nThen add 2 times \(\Delta H_{H_{2}(g) \to 2H(g)}\), to produce the required 2 moles of hydrogen.\nFinally, subtract 4 times \(\Delta H_{C-H bond}\). The negative sign is because this value represents bond breaking, which is an endothermic process, while we are interested in bond making, which is exothermic. \nThe sum gives the enthalpy change for our target process. For this sum, every enthalpy value is given per mole of methane formed. This covers our target reaction.
04

Calculate the standard enthalpy of formation

We now can put the above together to find the enthalpy change. Our formula: \(\Delta H_{C(s) \to CH4(g)} = \Delta H_{C(s) \to C(g)} + 2 \cdot \Delta H_{H_{2}(g) \to 2H(g)} - 4 \cdot \Delta H_{C-H bond} \). \nPlug in the values to obtain \(\Delta H_{C(s) \to CH4(g)} = 716 \, kJ/mol + 2 \cdot 0.5 \cdot 872.8 \, kJ/mol - 4 \cdot 414 \, kJ/mol.\) \nThe result simplifies to \(\Delta H_{C(s) \to CH4(g)} = 716 + 872.8 - 1656 = -67.2 \, kJ/mol.\)

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