Question

Hydrogen gas is being considered as a fuel for automobiles. There are many chemical means for producing hydrogen gas from water. One of these reactions is $$\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$. In this case the form of carbon used is graphite. a. Calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction using data from Appendix 4. b. At what temperature is \(\Delta G^{\circ}=\) zero for this reaction? Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Short answer

The change in enthalpy (\(\Delta{H^{\circ}}\)) for the reaction is 131.3 kJ/mol and the change in entropy (\(\Delta{S^{\circ}}\)) is 133.82 J/(mol K). The temperature at which the change in Gibbs free energy (\(\Delta{G^{\circ}}\)) is zero for this reaction is 981 K.

Step-by-step solution

Calculate ∆H° for the reaction

Using the formula for the change in enthalpy, we need to find the enthalpy of formation for each component, which can be obtained from Appendix 4. From there, we have the following values: \(\Delta{fH^{\circ}}\)(C(graphite)) = 0 kJ/mol (since graphite is the standard state of carbon) \(\Delta{fH^{\circ}}\)(H_2O(g)) = -241.83 kJ/mol \(\Delta{fH^{\circ}}\)(CO(g)) = -110.53 kJ/mol \(\Delta{fH^{\circ}}\)(H_2(g)) = 0 kJ/mol (since hydrogen gas is the standard state of hydrogen) Now, we can calculate \(\Delta{H^{\circ}}\) using the formula: \(\Delta{H^{\circ}} = (1)(-110.53\,\mathrm{kJ/mol})+(1)(0\,\mathrm{kJ/mol})-(1)(0\,\mathrm{kJ/mol})-(1)(-241.83\,\mathrm{kJ/mol})=131.3\,kJ/mol.\)

Calculate ∆S° for the reaction

To calculate the change in entropy, we need the standard entropies for each component. These values can also be found in Appendix 4: \(S^{\circ}\)(C(graphite)) = 5.694 J/(mol K) \(S^{\circ}\)(H_2O(g)) = 188.840 J/(mol K) \(S^{\circ}\)(CO(g)) = 197.67 J/(mol K) \(S^{\circ}\)(H_2(g)) = 130.68 J/(mol K) Now, we can calculate \(\Delta{S^{\circ}}\) using the formula: \(\Delta{S^{\circ}} = (1)(197.67\,\mathrm{J/(mol\,K)})+(1)(130.68\,\mathrm{J/(mol\,K)})-(1)(5.694\,\mathrm{J/(mol\,K)})-(1)(188.840\,\mathrm{J/(mol\,K)})= 133.82\,J/(mol\,K).\)

Find the temperature at which ∆G° is zero

We will use the third formula to find the temperature at which the change in Gibbs free energy is zero: \(\Delta{G^{\circ}} = \Delta{H^{\circ}} - T\Delta{S^{\circ}}\) Since we want \(\Delta{G^{\circ}}\) to be zero, we can set the equation as follows: \(0 = 131.3\,\mathrm{kJ/mol} - T(133.82\,\mathrm{J/(mol\,K)})\) Now, solve for temperature (T): \(T = \frac{131.3\,\mathrm{kJ/mol}}{133.82\,\mathrm{J/(mol\,K)}}\) Converting kJ to J: \(T = \frac{131300\,\mathrm{J/mol}}{133.82\,\mathrm{J/(mol\,K)}} = 981\,\mathrm{K}\) So, the temperature at which \(\Delta{G^{\circ}}\) is zero for this reaction is 981 K. This temperature is important because it is the transition point at which the reaction becomes spontaneous or non-spontaneous.

Key concepts

Enthalpy of Reaction

The enthalpy of reaction, often represented by \( \Delta H \), is a measure of the heat content change during a chemical reaction. For a reaction occurring at constant pressure, this change is the heat absorbed or released. Imagine a group of students performing a chemical reaction in a lab setting where heat is either being given off or absorbed. This heat change they observe or measure is related to the enthalpy change.

In our exercise, calculating the enthalpy change involves using the standard enthalpies of formation of the reactants and products. This is achieved by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products. For the given reaction producing hydrogen gas, we have \(\Delta H^\circ = 131.3\, kJ/mol\), implying the reaction absorbs heat from the surroundings, since the value is positive.

Standard Enthalpy of Formation

The standard enthalpy of formation, \(\Delta fH^\circ\), is the change in enthalpy when one mole of a substance is formed from its elements in their most stable states under standard conditions (1 atm pressure and 298.15 K temperature).

For instance, in the combustion of the graphite form of carbon to produce carbon monoxide gas, the standard enthalpy of formation is considered. Substances in their standard state, like graphite for carbon and diatomic hydrogen gas for hydrogen, have a \(\Delta fH^\circ\) of zero. These reference points serve as baselines to determine the heat of reaction, as seen in the problem where the enthalpy of formation values from Appendix 4 are crucial to the calculations.

Gibbs Free Energy

Gibbs free energy, denoted as \(G\), is a thermodynamic quantity that combines enthalpy and entropy to predict the spontaneity of a process. The change in Gibbs free energy, \(\Delta G\), for a reaction can be calculated using the equation \(\Delta G = \Delta H - T\Delta S\), where \(T\) is the absolute temperature in Kelvin.

The equation is a balance between two competing aspects: enthalpy, which reflects the total heat content, and entropy, which represents the amount of disorder. When \(\Delta G\) is negative, the process is spontaneous, and energy is released. For the reaction in consideration, we've found the temperature at which \(\Delta G^\circ = 0\), pinpointing the threshold of spontaneity.

Entropy

Entropy, symbolized by \(S\), is a fundamental concept in thermodynamics and represents the measure of the disorder or randomness in a system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time.

To envision entropy, imagine a child's room getting gradually messier unless effort is invested in tidying it up, this gradual increase in 'messiness' parallels a system's increase in entropy. In our exercise, entropy change \(\Delta S^\circ\) represents the difference in disorder between the products and reactants. It can imply the energy distribution within a system. For the reaction to produce hydrogen gas, we've calculated an entropy change of \(133.82\, J/(mol\,K)\), which indicates an increase in systemic disorder.