Animated Solution for Chemistry - Chemical Thermodynamics: Given, $C_{\text{(graphite)}}+O_2(g)⟶C{O}_2(g)$; ${\Delta }_r{H}^{\circ }=-393.5{\text{ kJ mol}}^{-1}$ $H_2(g)+\frac{1}{2}{O}_2(g)⟶H_{2}O(l)$; ${\Delta }_r{H}^{\circ }=-285.8{\text{ kJ mol}}^{-1}$ $C{O}_2(g)+2H_{2}O(l)⟶C{H}_4(g)+2O_2(g)$; ${\Delta }_r{H}^{\circ }=+890.3{\text{ kJ mol}}^{-1}$ Based on the above thermochemical equations, the value of ${\Delta }_r{H}^{\circ }$ at 298 K for the reaction, $C_{\text{(graphite)}}+2H_2(g)⟶C{H}_4(g)$ will be

Imagine you are given a set of Lego blocks, but instead of building a spaceship, you are building a chemical reaction. In thermochemistry, we often know the energy changes for several common reactions, like combustion, but we want to find the energy change for a reaction that is difficult to measure directly.

This is where Hess's Law of Constant Heat Summation comes to our rescue. It states that the total enthalpy change for a chemical reaction is completely independent of the pathway or the number of steps taken to achieve it. As long as the initial reactants and final products are the same, the total energy change remains constant. This powerful principle allows us to treat chemical equations just like algebraic equations—we can add them, subtract them, and multiply them by constants.

Let's look at the three thermochemical equations provided in the problem:

1. The combustion of graphite to form carbon dioxide: $C_{\text{(graphite)}}+O_2(g)⟶C{O}_2(g)\Delta H_1^{\circ }=-393.5{\text{ kJ mol}}^{-1}$

2. The combustion of hydrogen gas to form liquid water: $H_2(g)+\frac{1}{2}{O}_2(g)⟶H_{2}O(l)\Delta H_2^{\circ }=-285.8{\text{ kJ mol}}^{-1}$

3. A reaction that looks like the reverse of methane combustion: $C{O}_2(g)+2H_{2}O(l)⟶C{H}_4(g)+2O_2(g)\Delta H_3^{\circ }=+890.3{\text{ kJ mol}}^{-1}$

Our ultimate goal is to find the standard enthalpy of formation of methane, which is represented by the target reaction: $C_{\text{(graphite)}}+2H_2(g)⟶C{H}_4(g)$

To build our target reaction, we need to manipulate the given equations so that when we add them together, everything except our desired reactants and products cancels out.

First, we need one mole of $C_{\text{(graphite)}}$ on the reactant side. Equation 1 already has exactly this, so we will keep Equation 1 exactly as it is.

Second, we need two moles of $H_2(g)$ on the reactant side. Equation 2 only has one mole of $H_2(g)$. Therefore, we must multiply the entire Equation 2 by a factor of 2. Crucially, according to Hess's Law, we must also multiply its enthalpy change by 2.

Finally, we need one mole of $C{H}_4(g)$ on the product side. Equation 3 already has this perfectly positioned, so we keep Equation 3 as it is.

Our master equation for the total enthalpy change becomes: $\Delta H_{\text{target}}^{\circ }=\Delta H_1^{\circ }+2\times \Delta H_2^{\circ }+\Delta H_3^{\circ }$

Now, let's substitute the raw numerical values into our master equation. Don't rush through this; watch out for the minus signs!

First, multiply the enthalpy of the second reaction: $\Delta H_{\text{target}}^{\circ }=-393.5-571.6+890.3$

Next, combine the negative terms: $\Delta H_{\text{target}}^{\circ }=-965.1+890.3$

Finally, add the positive term to find the net enthalpy change: $\Delta H_{\text{target}}^{\circ }=-74.8{\text{ kJ mol}}^{-1}$

This negative value indicates that the formation of methane from its constituent elements is an exothermic process. The result perfectly matches option (c).

Alternative Insight: You could also solve this by recognizing that the first two equations directly provide the standard enthalpies of formation for $C{O}_2$ and $H_{2}O$. By plugging these values into the standard formula ${\Delta }_r{H}^{\circ }=\sum {\Delta }_f{H}_{\text{products}}^{\circ }-\sum {\Delta }_f{H}_{\text{reactants}}^{\circ }$ for the third equation, you can algebraically solve for the unknown ${\Delta }_f{H}^{\circ }$ of $C{H}_4$. Both paths lead to the exact same beautiful result!