Question

Calculate \(\Delta G_{R}^{\circ}\) for the reaction \(\mathrm{CO}(g)+\) \(1 / 2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)\) at \(298.15 \mathrm{K} .\) Calculate \(\Delta G_{R}^{\circ}\) at \(600 . \mathrm{K}\) assuming that \(\Delta H_{R}^{\circ}\) is constant in the temperature interval of interest.

Short answer

The standard Gibbs free energy change (\(\Delta G_{R}^{\circ}\)) for the given reaction CO(g) + 1/2 O₂(g) → CO₂(g) is approximately -257.19 kJ/mol at 298.15 K and -209.06 kJ/mol at 600.00 K.

Step-by-step solution

Calculate \(\Delta G_{R}^{\circ}\) at 298.15 K

To calculate \(\Delta G_{R}^{\circ}\) for the reaction at 298.15 K, we will use the following equation: \(\Delta G_{R}^{\circ} = \Delta G_{f,CO₂}^{\circ} - \Delta G_{f,CO}^{\circ} - \frac{1}{2}\Delta G_{f,O₂}^{\circ}\) We need to determine the standard Gibbs free energies of formation (\(\Delta G_{f}^{\circ}\)) for the species involved in the reaction. These values can be found in thermodynamic tables: \(\Delta G_{f,CO}^{\circ}(298.15 K) = -137.17 \, \text{kJ/mol}\) \(\Delta G_{f,CO₂}^{\circ} (298.15 K)= -394.36 \, \text{kJ/mol}\) \(\Delta G_{f,O₂}^{\circ} (298.15 K)= 0 \, \text{kJ/mol}\) (as it is an element in its standard state) Now, using these values, we can calculate \(\Delta G_{R}^{\circ}\) for the reaction at 298.15 K: \(\Delta G_{R}^{\circ} (-394.36) - (-137.17) - \frac{1}{2}(0) = -257.19 \, \text{kJ/mol}\)

Calculate \(\Delta G_{R}^{\circ}\) at 600.00 K

To calculate \(\Delta G_{R}^{\circ}\) at 600 K, we will use the Gibbs-Helmholtz equation: \(\Delta G_{R}^{\circ}(T) = \Delta G_{R}^{\circ}(298.15 \, \text{K}) + \Delta H_{R}^{\circ}(298.15 \, \text{K})\cdot\left(\frac{1}{T} - \frac{1}{298.15 \, \text{K}}\right)\) Now, we need to find the standard enthalpy change of the reaction (\(\Delta H_{R}^{\circ}\)) at 298.15 K. We can use the standard enthalpies of formation (\(\Delta H_{f}^{\circ}\)) of the species involved in the reaction: \(\Delta H_{R}^{\circ}(298.15 \, \text{K}) = \Delta H_{f,CO₂}^{\circ}(298.15 \, \text{K}) - \Delta H_{f,CO}^{\circ}(298.15 \, \text{K}) - \frac{1}{2}\Delta H_{f,O₂}^{\circ}(298.15 \, \text{K})\) Again, we can find these values in thermodynamic tables: \(\Delta H_{f,CO}^{\circ} (298.15\, \text{K}) = -110.53 \, \text{kJ/mol}\) \(\Delta H_{f,CO₂}^{\circ} (298.15\, \text{K}) = -393.51 \, \text{kJ/mol}\) \(\Delta H_{f,O₂}^{\circ} (298.15\, \text{K}) = 0 \, \text{kJ/mol}\) (as it is an element in its standard state) Now we can calculate the standard enthalpy change of the reaction: \(\Delta H_{R}^{\circ}(298.15 \, \text{K}) = (-393.51) - (-110.53) - \frac{1}{2}(0) = -282.98 \, \text{kJ/mol}\) Now, we can use the Gibbs-Helmholtz equation to calculate \(\Delta G_{R}^{\circ}\) at 600.00 K: \(\Delta G_{R}^{\circ}(600\, \text{K}) = -257.19 \, \text{kJ/mol} - 282.98 \, \text{kJ/mol}\cdot\left(\frac{1}{600\, \text{K}} - \frac{1}{298.15\, \text{K}}\right)\) \(\Delta G_{R}^{\circ}(600\, \text{K}) \approx -209.06 \, \text{kJ/mol}\) So, the standard Gibbs free energy change (\(\Delta G_{R}^{\circ}\)) for the given reaction is approximately -257.19 kJ/mol at 298.15 K and -209.06 kJ/mol at 600.00 K.

Key concepts

Standard Gibbs Free Energy Change

The standard Gibbs free energy change, denoted as \( \Delta G^{\circ} \), serves as a powerful tool in thermodynamics to predict whether a reaction will proceed spontaneously under standard conditions.This property considers both enthalpy and entropy changes, essential variables defining chemical processes.
Standard conditions are set at 298.15 K (25°C) and 1 atm pressure, and the reference state of each element is defined as its naturally occurring form.For reactions, the calculation of \( \Delta G^{\circ} \) involves summing the Gibbs free energies of formation of products and subtracting those of reactants.
In our given exercise, the equation used was:\[ \Delta G_{R}^{\circ} = \Delta G_{f,CO₂}^{\circ} - \Delta G_{f,CO}^{\circ} - \frac{1}{2}\Delta G_{f,O₂}^{\circ} \]The reaction involves carbon monoxide, oxygen, and carbon dioxide.By substituting their standard free energies of formation, we determined \( \Delta G_{R}^{\circ} = -257.19 \, \text{kJ/mol} \) at 298.15 K.A negative \( \Delta G^{\circ} \) signifies that the reaction proceeds spontaneously under these conditions.

Gibbs-Helmholtz Equation

The Gibbs-Helmholtz Equation is a crucial component in thermodynamics for exploring how the Gibbs free energy changes with temperature.It's particularly beneficial when calculating \( \Delta G^{\circ} \) at temperatures different from the standard state.
The Gibbs-Helmholtz equation is expressed as:\[ \Delta G_{R}^{\circ}(T) = \Delta G_{R}^{\circ}(298.15 \text{ K}) + \Delta H_{R}^{\circ}(298.15 \text{ K}) \left(\frac{1}{T} - \frac{1}{298.15 \text{ K}}\right) \]Here, \( \Delta H_{R}^{\circ} \) represents the standard enthalpy change, assumed constant over the temperature range in question.
Utilizing this equation, our exercise progressed to calculate \( \Delta G_{R}^{\circ} \) at 600 K.We estimated a decreased free energy value of approximately -209.06 kJ/mol, highlighting that the reaction is still spontaneous, albeit slightly less favorable than at standard conditions.

Standard Enthalpy Change

The standard enthalpy change, \( \Delta H^{\circ} \), represents the heat absorbed or released during a reaction under standard conditions.This concept is significant as it helps evaluate whether a reaction is exothermic or endothermic.
Calculating \( \Delta H_{R}^{\circ} \) involves the enthalpies of formation of the reactants and products:\[ \Delta H_{R}^{\circ} = \Delta H_{f,CO₂}^{\circ} - \Delta H_{f,CO}^{\circ} - \frac{1}{2}\Delta H_{f,O₂}^{\circ} \]For our exercise, enthalpies of formation were retrieved from thermodynamic tables, yielding \( \Delta H_{R}^{\circ} = -282.98 \text{ kJ/mol} \).
This negative \( \Delta H_{R}^{\circ} \) value confirms that the reaction releases heat, thus being exothermic.These insights are instrumental when linked with the Gibbs-Helmholtz Equation to deduce the impact of temperature changes on \( \Delta G^{\circ} \), as observed in calculations at 600 K.