Question

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

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

Step-by-step solution

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\( \mathrm{C}(s) + 2\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g) \)

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.

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.

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 \(\Delta H_{C(s) \to CH4(g)} = 716 + 872.8 - 1656 = -67.2 \, kJ/mol.\)

Key concepts

Hess's Law

Hess's Law is a very useful tool in thermodynamics and chemistry because it allows us to calculate the energy changes in a chemical reaction. It's based on the principle that the total enthalpy change of a reaction is the same, regardless of the pathway taken, as long as the starting and ending conditions are the same. This means whether a reaction happens in one step or multiple steps, the overall change in enthalpy remains constant.

Applying Hess's Law involves breaking down a complex reaction into simpler ones for which enthalpies are known. We then add these enthalpies algebraically to determine the overall enthalpy change for the target reaction. For example, when calculating the standard enthalpy of formation of methane, we decomposed the process into converting carbon and hydrogen to their gaseous states and forming the C-H bonds. By summing these known values, we use Hess's Law to find our answer.

• Hess's Law is essential when direct measurement of a reaction's enthalpy change is difficult.
• It involves adding enthalpy changes of individual steps.
• Reacts can be multiplied or flipped to match the target equation, changing their enthalpy values accordingly.

Bond Enthalpy

Bond enthalpy, also known as bond dissociation energy, is the energy required to break a particular chemical bond in a mole of gaseous molecules. It's crucial because it gives insight into the strength of a bond: the higher the bond enthalpy, the stronger the bond. For instance, the C-H bond in methane has an average bond enthalpy of 414 kJ/mol, indicating considerable energy is required to break each bond.

In reactions, breaking bonds requires energy, making such steps endothermic, while forming bonds releases energy, an exothermic process. When formations involve multiple similar bonds, like in methane's four C-H bonds, we can multiply the average bond enthalpy by the number of bonds, which helps in calculating total enthalpy change.

• Breaking bonds takes energy, meaning endothermic steps.
• Forming bonds releases energy, meaning exothermic steps.
• A bond enthalpy can be used to estimate reaction energy changes, especially when specific formation enthalpies aren't available.

Standard Enthalpy Change

The standard enthalpy change is an essential concept in understanding how much heat energy is absorbed or released during a chemical reaction under standard conditions, which include a pressure of 1 bar and substances in their standard states. Calculating the standard enthalpy change, specifically the enthalpy of formation, involves summing the enthalpy changes of reactions leading to the formation of a compound from its elements in their most stable form. For methane's formation, it is represented as converting solid carbon and gaseous hydrogen into gaseous methane, under standard conditions.

Standard Enthalpy of Formation - Defined as the enthalpy change when one mole of a compound forms from its elements in their standard states. - Commonly used to compute enthalpy changes in reactions when direct data isn't available. When working with standard enthalpy changes, using known values, like bond enthalpies and Hess's Law, allows us to indirectly find values for reactions of interest.