Question

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the controlled oxidation of methane: $$ \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. (b) How is \(\Delta G^{\circ}\) for the reaction expected to vary with increasing temperature? (c) Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable?

Short answer

The change in enthalpy (∆H) for the reaction is -126.39 kJ/mol, and the change in entropy (∆S) is -69.05 J/mol•K. The Gibbs Free Energy (∆G) increases with increasing temperature. At 298 K, ∆G is -105.8061 kJ/mol, and the reaction is spontaneous under standard conditions. The equilibrium temperature is approximately 1829.93 K, but this value is too high for Methane, Oxygen, and Methanol to be stable simultaneously.

Step-by-step solution

(a) Calculating ∆H and ∆S for this reaction

To calculate ∆H and ∆S, we must first find the values for Methane, Oxygen and Methanol involved in the reaction from Appendix C. Enthalpy (H) for the reactants and products are: - CH₄(g): -74.81 kJ/mol - O₂(g): 0 kJ/mol - CH₃OH(g): -201.2 kJ/mol Entropy (S) for the reactants and products are: - CH₄(g): 186.3 J/mol•K - O₂(g): 205.2 J/mol•K - CH₃OH(g): 239.7 J/mol•K Using the values, we can now calculate ∆H and ∆S as follows: ∆H = ∑ H(products) - ∑ H(reactants) ∆H = (-201.2 kJ/mol) - [(-74.81 kJ/mol) + (0 kJ/mol)] ∆H = -126.39 kJ/mol ∆S = ∑ S(products) - ∑ S(reactants) ∆S = (239.7 J/mol•K) - [(186.3 J/mol•K) + (1/2)(205.2 J/mol•K)] ∆S = -69.05 J/mol•K So, ∆H = -126.39 kJ/mol and ∆S = -69.05 J/mol•K

(b) Variation of ∆G with increasing temperature

The Gibbs Free Energy change (∆G) can be calculated using the equation: ∆G = ∆H - T∆S As we can see, the change in Gibbs Free Energy is directly dependent on the temperature (T). Since ∆S is negative, as the temperature increases, the value of -T∆S increases, which would result in an increase in ∆G.

(c) Calculate ∆G at 298 K and its spontaneity

Using the earlier equation (∆G = ∆H - T∆S), we can now calculate ∆G for the given temperature (298 K): ∆G = (-126.39 kJ/mol) - (298 K * (-69.05 J/mol•K)) Convert J to kJ: -69.05 J/mol•K * (1 kJ / 1000 J) = -0.06905 kJ/mol•K ∆G = (-126.39 kJ/mol) - (298 K * (-0.06905 kJ/mol•K)) ∆G = (-126.39 kJ/mol) + 20.5839 kJ/mol ∆G = -105.8061 kJ/mol Since ∆G is negative at 298 K, the reaction is spontaneous under standard conditions.

(d) Finding the equilibrium temperature

To find the temperature where the reaction is at equilibrium, we can use the equilibrium condition: ∆G = 0 From the earlier equation (∆G = ∆H - T∆S), we can derive the temperature as follows: 0 = -126.39 kJ/mol - T*(-0.06905 kJ/mol•K) T = -126.39 kJ/mol / -0.06905 kJ/mol•K T = 1829.93 K The equilibrium temperature is approximately 1829.93 K. However, this temperature will likely cause Methane, Oxygen, and Methanol to break down into various other compounds due to the high temperature. Therefore, it is unlikely that there is a temperature low enough for these compounds to be stable and reach equilibrium simultaneously.

Key concepts

Enthalpy Change Calculation

Understanding the enthalpy change of a chemical reaction is crucial for students studying chemical thermodynamics. The enthalpy change, often denoted as \(\Delta H\), represents the heat absorbed or released during a reaction at constant pressure. It's a state function, implying that its value depends only on the initial and final states of the reaction, not the path taken to get from one to the other.

To calculate \(\Delta H\), one must know the enthalpy values of the reactants and products involved in the reaction. These values are usually provided in textbooks or reference material like Appendix C in the given exercise. The enthalpy change can be found using the formula:
\[\Delta H = \sum H(\text{products}) - \sum H(\text{reactants})\]
This calculation yields the enthalpy change for the formation of Methanol from Methane and Oxygen. A negative \(\Delta H\) value, as seen in the exercise, indicates an exothermic reaction—meaning heat is released in the process.

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) is a thermodynamic quantity that combines the system's enthalpy and entropy to determine the spontaneity of a chemical reaction. At a constant temperature and pressure, the change in Gibbs Free Energy provides insight into whether a reaction will occur spontaneously. The Gibbs Free Energy equation is:
\[\Delta G = \Delta H - T\Delta S\]
where \(T\) represents the absolute temperature in Kelvin (K), and \(\Delta S\) is the change in entropy. A negative value of \(\Delta G\) indicates a spontaneous process, whereas a positive value implies non-spontaneity. In the context of the provided exercise, the calculation of \(\Delta G\) at a specific temperature elucidates the reaction's feasibility at that given state.

Spontaneous Reactions

A spontaneous reaction is one that occurs without the need for a continuous external input of energy. The spontaneity of a reaction can be understood in terms of Gibbs Free Energy. If the Gibbs Free Energy change \(\Delta G\) is negative, the reaction is spontaneous under the given conditions, meaning it will proceed in the forward direction on its own. Conversely, if \(\Delta G\) is positive, the reaction is non-spontaneous and will not occur without external energy. The temperature and entropy change both significantly influence the spontaneity. As temperature increases, the negative entropy change (\(\Delta S\)) can become more influential, potentially converting a non-spontaneous reaction into a spontaneous one at higher temperatures.

For example, in the textbook exercise, the reaction is found to be spontaneous at 298 K since the calculated \(\Delta G\) is negative.

Chemical Equilibrium Temperature

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. The temperature at which a reaction reaches equilibrium is known as the chemical equilibrium temperature. At this point, Gibbs Free Energy \(\Delta G\) is zero. Using the formula provided in the exercise, the equilibrium temperature can be found when \(\Delta H\) and \(\Delta S\) are known.
\[0 = \Delta H - T_{eq}\Delta S\]
Solving this equation for \(T_{eq}\) gives the equilibrium temperature. However, this temperature must be within a stable range for the substances involved to ensure that they don't decompose or otherwise change states. In the exercise, the equilibrium temperature is calculated to be 1829.93 K, but this is likely too high to maintain the stability of Methane, Oxygen, and Methanol.