Today, most candles are made of paraffin wax. A typical component of paraffin wax is the hydrocarbon \(\mathrm{C}_{31} \mathrm{H}_{64}\) which is solid at room temperature. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{31} \mathrm{H}_{64}(s)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) ( \(\mathbf{b}\) ) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

To write a balanced chemical equation, first write down the basic form of the equation with the reactants on the left side and the products on the right side. Then, we will balance the equation by adjusting the number of each molecule to make sure the number of each atom is the same on both sides of the equation. The basic form of the equation is: \(\mathrm{C}_{31}\mathrm{H}_{64}(s) + O_2(g) \rightarrow CO_2(g) + H_2O(g)\) Now, let's balance the equation: There are 31 carbon atoms in \(\mathrm{C}_{31}\mathrm{H}_{64}\), so we need 31 molecules of \(\mathrm{CO}_2\) on the right side of the equation: \(\mathrm{C}_{31}\mathrm{H}_{64}(sockopt) + O_2(g) \rightarrow \textbf{31} CO_2(g) + H_2O(g)\) Now, there are 64 hydrogen atoms in \(\mathrm{C}_{31}\mathrm{H}_{64}\), so we need 32 molecules of \(H_2O\) on the right side of the equation: \(\mathrm{C}_{31}\mathrm{H}_{64}(s) + O_2(g) \rightarrow 31 CO_2(g) + \textbf{32} H_2O(g)\) Finally, we have 31 molecules of carbon dioxide (31x2=62 oxygen atoms) and 32 molecules of water vapor (32x1=32 oxygen atoms), which combine for a total of 94 oxygen atoms needed on the left side of the equation. Thus, we need 47 molecules of \(O_2\): \(\mathrm{C}_{31}\mathrm{H}_{64}(s) + \textbf{47} O_2(g) \rightarrow 31 CO_2(g) + 32 H_2O(g)\) Now, the balanced chemical equation is: \[ \mathrm{C}_{31}\mathrm{H}_{64}(s) + 47 O_2(g) \rightarrow 31 CO_2(g) + 32 H_2O(g) \]

First, let's recall the Gibbs free energy equation: \[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \] Since we can't use thermochemical data, we must make a prediction based on other factors. Combustion reactions are typically exothermic, which means that they release heat. As a result, the standard enthalpy change \(\Delta H^{\circ}\) will be negative. Additionally, since we are going from a relatively ordered hydrocarbon solid and oxygen gas to a more chaotic mixture of gases, carbon dioxide and water vapor, the entropy change \(\Delta S^{\circ}\) will be positive. Considering the negative sign in the Gibbs free energy equation, the product of positive \(\Delta S^{\circ}\) and a positive temperature (T) will also be positive, which means the whole term \(-T \Delta S^{\circ}\) will be negative. Since both terms of the right side of the Gibbs free energy equation are negative, \(\Delta G^{\circ}\) will be more negative than \(\Delta H^{\circ}\). So, without using thermochemical data, we predict that \(\Delta G^{\circ}\) for this reaction is more negative than \(\Delta H^{\circ}\).