Import Question JSON

$\text{Hint: Enthalpy of formation of the compound can be positive or negative depending on the type of reaction.}$ $\text{Explanation:}$ $\text{In a chemical reaction, the reactant first converts into its atomic state, then combined, and formed a product.}$ $\text{In the first reaction, C and H are present in their atomic state hence they will combine and release x amount of energy.}$ $\text{In the second reaction, H}_2 \text{ molecule first converts into atomic hydrogen and Graphite first converts into Carbon(g) then combined with carbon atom and formed a product.}$ $\text{The overall } \Delta H_r \text{ for both reactions are as follows:}$ $i. \Delta H_r = x = -\text{energy of the bonds being formed (four C-H bonds)}$ $ii. \Delta H_r = y = \text{energy of the bonds being broken (H-H bonds)} - \text{energy of the bonds being formed (four C-H bonds)}$ $\text{In both reactions, the energy of the bonds being formed is equal but in the second reaction, extra energy is used to break the hydrogen molecule bonds and to convert graphite to gaseous carbon.}$ $\text{To analyze this more formally:}$ $\text{For reaction (i), we directly form CH}_4 \text{ from gaseous atoms:}$ $\text{C}(g) + 4\text{H}(g) \rightarrow \text{CH}_4(g); \Delta H_r = x$ $\text{For reaction (ii), we need to consider the atomization of the reactants:}$ $\text{C(graphite)} \rightarrow \text{C}(g); \Delta H_{\text{atomization}} > 0 \text{ (endothermic)}$ $\text{H}_2(g) \rightarrow 2\text{H}(g); \Delta H_{\text{dissociation}} > 0 \text{ (endothermic)}$ $\text{Since breaking bonds requires energy and the second reaction starts with more stable forms (graphite instead of gaseous carbon, and H}_2 \text{ molecules instead of H atoms), more energy needs to be input before bonds can form.}$ $\text{Therefore, the energy released in the first reaction (x) will be greater in magnitude than the energy released in the second reaction (y).}$ $\text{Thus, } x > y \text{.}$