Problem 67 - Chemistry: An Atoms-Focused Approach

Question

One method for the industrial production of methanol uses the following reaction: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\ell)$$ a. Use the data in Appendix 4 to calculate \(\Delta G^{\circ}\) for this reaction at \(298 \mathrm{K}\) b. The reaction is normally run at a minimum temperature of \(475 \mathrm{K}\). What is the value of \(\Delta G\) at that temperature? Is the reaction spontaneous at that temperature?

Step-by-Step Solution

Answer

In summary, the Gibbs free energy change (\(\Delta G^{\circ}\)) for the reaction at standard conditions (298 K) is -24.93 kJ/mol, and the Gibbs free energy change (\(\Delta G\)) at 475 K is approximately -20.05 kJ/mol. As both values are negative, the reaction is spontaneous at both temperatures.

Step 1: Calculate the Gibbs free energy change at standard conditions

Using the balanced equation, we can write the expression for the Gibbs free energy change at standard conditions as: $$\Delta G^{\circ} = \Delta G_f^{\circ}\text{(product)} - \Delta G_f^{\circ}\text{(reactants)}$$ We will now look up the standard state Gibbs free energy of formation values for each substance in Appendix 4. For CO: \(\Delta G_f^{\circ}(\mathrm{CO}) = -137.27\,\text{kJ/mol}\) For H2: \(\Delta G_f^{\circ}(\mathrm{H2}) = 0\,\text{kJ/mol}\) For CH3OH: \(\Delta G_f^{\circ}(\mathrm{CH3OH}) = -162.2\,\text{kJ/mol}\) Now, we can calculate \(\Delta G^{\circ}\) using these values: $$\Delta G^{\circ} = (1)(-162.2\,\text{kJ/mol}) - [(1)(-137.27\,\text{kJ/mol}) + (2)(0\,\text{kJ/mol})]$$ Calculating this, we get: $$\Delta G^{\circ} = -24.93\,\text{kJ/mol}$$

Step 2: Calculate the Gibbs free energy change at 475 K

Now, we can use the van't Hoff equation to determine the Gibbs free energy change at the non-standard temperature of \(T = 475\,\text{K}\). The equation is given by: $$\ln\left(\frac{\Delta G}{\Delta G^{\circ}}\right) = -\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T}-\frac{1}{298\,\text{K}}\right)$$ Since we do not have the value of \(\Delta H^{\circ}\), we will solve this problem slightly differently. We can use the relationship between \(\Delta G\), \(\Delta H\), and \(\Delta S\): $$\Delta G = \Delta H - T\Delta S$$ From this expression, we can infer that: $$\frac{\Delta G}{\Delta G^{\circ}} = \frac{\Delta H - T\Delta S}{\Delta H^{\circ} - 298\,\text{K}\Delta S^{\circ}}$$ Now we can use the van't Hoff equation: $$\ln\left(\frac{\Delta G}{\Delta G^{\circ}}\right) = \frac{\Delta S^{\circ}}{R}\left(\frac{298\,\text{K}T}{T - 298\,\text{K}}\right)$$ We will now look up the standard state entropy change values \((\Delta S^{\circ})\) for each substance in Appendix 4. For CO: \(\Delta S^{\circ}(\mathrm{CO}) = 197.7\,\text{J/mol} \cdot \text{K}\) For H2: \(\Delta S^{\circ}(\mathrm{H2}) = 130.7\,\text{J/mol} \cdot \text{K}\) For CH3OH: \(\Delta S^{\circ}(\mathrm{CH3OH}) = 251.6\,\text{J/mol} \cdot \text{K}\) Now, we can find the standard entropy change for the reaction as: $$\Delta S^{\circ} = \Delta S^{\circ}\text{(product)} - \Delta S^{\circ}\text{(reactants)}$$ We get: $$\Delta S^{\circ} = (251.6\,\text{J/mol} \cdot \text{K}) - [(197.7\,\text{J/mol} \cdot \text{K}) + (2)(130.7\,\text{J/mol} \cdot \text{K})] = -207.5\,\text{J/mol} \cdot \text{K}$$ Now, we can plug in the values of \(\Delta S^{\circ}\) and \(\Delta G^{\circ}\) into the modified van't Hoff equation and solve for \(\Delta G\) at 475 K: $$\ln\left(\frac{\Delta G}{-24.93\,\text{kJ/mol}}\right) = \frac{-207.5\,\text{J/mol} \cdot \text{K}}{8.3145\,\text{J/mol} \cdot \text{K}}\left(\frac{298\,\text{K}(475\,\text{K})}{475\,\text{K} - 298\,\text{K}}\right)$$ $$\Delta G = -24.93\,\text{kJ/mol} \times e^{\left(-207.5\,\text{J/mol} \cdot \text{K}\cdot\frac{298\,\text{K}(475\,\text{K})}{8.3145\,\text{J/mol} \cdot \text{K}(475\,\text{K} - 298\,\text{K})}\right)}$$ Calculating this, we get: $$\Delta G \approx -20.05\,\text{kJ/mol}$$

Step 3: Determine if the reaction is spontaneous

It is now clear that the value of \(\Delta G\) at the higher temperature of \(475\,\text{K}\) is negative: $$\Delta G \approx -20.05\,\text{kJ/mol}$$ Since the Gibbs free energy change is negative, we can conclude that the reaction is spontaneous at this temperature.

Key Concepts

Enthalpy Change

In thermodynamics, enthalpy change (\( \Delta H \)) is a central concept that helps us understand energy exchange in chemical reactions. Enthalpy is essentially the heat content of a system and is used to measure the total change in energy when reactants are converted to products.
This change can tell us whether heat is absorbed or released during the reaction:

• If \( \Delta H \) is negative, the reaction releases heat and is exothermic.
• If \( \Delta H \) is positive, the reaction absorbs heat and is endothermic.

When calculating the Gibbs free energy (\( \Delta G \)), the enthalpy change is crucial as it directly affects reaction spontaneity.
Analyzing the reaction of methanol production, knowing the enthalpy change would help understand how temperature and other conditions influence the process.

Entropy Change

Entropy change (\( \Delta S \)) is another important factor in reaction thermodynamics. Entropy is a measure of the disorder or randomness in a system. When a chemical reaction occurs, the entropy of the system can increase or decrease.

• An increase in entropy (\( \Delta S \) > 0 ) suggests greater disorder, which is generally favorable since nature tends toward disorder.
• A decrease in entropy (\( \Delta S \) < 0 ) implies a reduction in disorder.

Entropy change is significant because it, along with enthalpy change, influences the Gibbs free energy of a system:
\[\Delta G = \Delta H - T\Delta S\]For methanol production, the calculated \( \Delta S \) was found to be negative, indicating decreasing disorder. Understanding this will help when predicting how temperature changes impact reaction outcomes.

Reaction Spontaneity

Reaction spontaneity is determined by the Gibbs free energy change (\( \Delta G \)) of a reaction. A reaction is spontaneous if \( \Delta G \) is negative, meaning the process can occur without external input of energy.
Let's break down what contributes to this crucial parameter:

• If the enthalpy change (\( \Delta H \) ) is negative and entropy change (\( \Delta S \) ) is positive, reactions are more likely to be spontaneous.
• The temperature also plays a role, as seen in the \( \Delta G = \Delta H - T\Delta S \) formula.

In the methanol formation reaction, we calculated \( \Delta G \) at two different temperatures: standard 298 K and elevated 475 K. At both temperatures, \( \Delta G \) was negative, indicating that the reaction spontaneously favors the formation of methanol. This confirms the industrial reaction's efficiency under variable temperatures and helps in optimizing production strategies.