Problem 55
Question
Predict whether the reaction given is product-favored or reactant-favored by calculating \(\Delta_{\mathrm{r}} G^{\circ}\) from the entropy and enthalpy changes for the reaction at \(25^{\circ} \mathrm{C}\). \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) $$ \Delta_{r} H^{\circ}=41.17 \mathrm{~kJ} / \mathrm{mol} \quad \Delta_{r} S^{\circ}=42.08 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} $$
Step-by-Step Solution
Verified Answer
The reaction is reactant-favored.
1Step 1: Understand the Given Data
We are given the reaction and need to determine if it is product-favored or reactant-favored by calculating the standard Gibbs free energy change, \( \Delta_{\mathrm{r}} G^{\circ} \). We have the values for \( \Delta_{\mathrm{r}} H^{\circ} \) (enthalpy change) and \( \Delta_{\mathrm{r}} S^{\circ} \) (entropy change) at \( 25^{\circ} \mathrm{C} \) which is \( 298 \mathrm{~K} \).
2Step 2: Convert Units
Ensure all units are consistent. \( \Delta_{\mathrm{r}} H^{\circ} = 41.17 \mathrm{~kJ/mol} \) needs to be converted to \( \mathrm{J/mol} \) to match \( \Delta_{\mathrm{r}} S^{\circ} \). Thus, \( \Delta_{\mathrm{r}} H^{\circ} = 41170 \mathrm{~J/mol} \).
3Step 3: Apply the Gibbs Free Energy Formula
Use the equation:\[\Delta_{\mathrm{r}} G^{\circ} = \Delta_{\mathrm{r}} H^{\circ} - T \Delta_{\mathrm{r}} S^{\circ}\]Substitute the values: \( \Delta_{\mathrm{r}} H^{\circ} = 41170 \mathrm{~J/mol} \), \( T = 298 \mathrm{~K} \), and \( \Delta_{\mathrm{r}} S^{\circ} = 42.08 \mathrm{~J/K \cdot mol} \).
4Step 4: Calculate \( \Delta_{\mathrm{r}} G^{\circ} \)
Perform the calculation:\[\Delta_{\mathrm{r}} G^{\circ} = 41170 \mathrm{~J/mol} - (298 \mathrm{~K} \times 42.08 \mathrm{~J/K \cdot mol})\]\[\Delta_{\mathrm{r}} G^{\circ} = 41170 \mathrm{~J/mol} - 12529.84 \mathrm{~J/mol}\]\[\Delta_{\mathrm{r}} G^{\circ} = 28640.16 \mathrm{~J/mol}\]
5Step 5: Determine Reaction Favorability
The value of \( \Delta_{\mathrm{r}} G^{\circ} = 28640.16 \mathrm{~J/mol} \) is positive, indicating that the reaction is reactant-favored at \( 25^{\circ} \mathrm{C} \).
Key Concepts
Enthalpy ChangeEntropy ChangeReaction Favorability
Enthalpy Change
Enthalpy change, denoted as \( \Delta_{r} H^{\circ} \), represents the heat absorbed or released during a chemical reaction at constant pressure. It is essentially a measure of the energy change in the form of heat. In the context of our reaction, \( \Delta_{r} H^{\circ} = 41.17 \, \text{kJ/mol} \). This value tells us that the reaction is endothermic, meaning it absorbs heat from the surroundings.
Enthalpy is measured in kilojoules per mole (\( \text{kJ/mol} \)), but for some calculations, like Gibbs Free Energy, we need to convert it to joules per mole (\( \text{J/mol} \)) by multiplying by 1000. Thus, \( \Delta_{r} H^{\circ} = 41170 \, \text{J/mol} \).
When analyzing reactions, knowing the enthalpy change helps predict how much heat is needed for the reaction to proceed and aids in understanding whether a reaction will require energy (endothermic) or release energy (exothermic).
Enthalpy is measured in kilojoules per mole (\( \text{kJ/mol} \)), but for some calculations, like Gibbs Free Energy, we need to convert it to joules per mole (\( \text{J/mol} \)) by multiplying by 1000. Thus, \( \Delta_{r} H^{\circ} = 41170 \, \text{J/mol} \).
When analyzing reactions, knowing the enthalpy change helps predict how much heat is needed for the reaction to proceed and aids in understanding whether a reaction will require energy (endothermic) or release energy (exothermic).
Entropy Change
Entropy change, represented by \( \Delta_{r} S^{\circ} \), signifies the disorder or randomness in a system. It's an important factor because it helps determine the spontaneity of a reaction. For our reaction, \( \Delta_{r} S^{\circ} = 42.08 \, \text{J/K} \cdot \text{mol} \), indicating a slight increase in disorder due to the products formed.
Understanding entropy helps chemists predict how the distribution of energy will shift during a chemical process. A positive entropy change suggests that the reaction results in increased disorder, which is often favorable, depending on temperature conditions.
In the grand scheme of a chemical reaction, even a small change in entropy can significantly impact whether a reaction proceeds spontaneously, so it is crucial to consider together with enthalpy and temperature effects.
Understanding entropy helps chemists predict how the distribution of energy will shift during a chemical process. A positive entropy change suggests that the reaction results in increased disorder, which is often favorable, depending on temperature conditions.
In the grand scheme of a chemical reaction, even a small change in entropy can significantly impact whether a reaction proceeds spontaneously, so it is crucial to consider together with enthalpy and temperature effects.
Reaction Favorability
Reaction favorability determines whether a reaction is likely to proceed in forming products or revert to reactants. The key metric for this assessment is the Gibbs Free Energy change, \( \Delta_{r} G^{\circ} \). Calculated using the equation \( \Delta_{r} G^{\circ} = \Delta_{r} H^{\circ} - T \Delta_{r} S^{\circ} \), it combines the effects of enthalpy and entropy at a given temperature, \( T \).
For our specific reaction at \( 25^\circ C \) or \( 298 \, \text{K} \), the calculated \( \Delta_{r} G^{\circ} = 28640.16 \, \text{J/mol} \). Since this value is positive, it suggests the reaction is reactant-favored under these conditions, meaning it does not spontaneously proceed to products.
A negative \( \Delta_{r} G^{\circ} \) would indicate that the reaction is product-favored and spontaneous, whereas a positive value like in our example requires conditions to shift for spontaneity. Understanding this balance aids in manipulating conditions for desired chemical processes, such as increasing temperature to potentially favor the formation of products if entropy plays a greater role at higher thermal energy.
For our specific reaction at \( 25^\circ C \) or \( 298 \, \text{K} \), the calculated \( \Delta_{r} G^{\circ} = 28640.16 \, \text{J/mol} \). Since this value is positive, it suggests the reaction is reactant-favored under these conditions, meaning it does not spontaneously proceed to products.
A negative \( \Delta_{r} G^{\circ} \) would indicate that the reaction is product-favored and spontaneous, whereas a positive value like in our example requires conditions to shift for spontaneity. Understanding this balance aids in manipulating conditions for desired chemical processes, such as increasing temperature to potentially favor the formation of products if entropy plays a greater role at higher thermal energy.
- Negative \( \Delta_{r} G^{\circ} \): Reaction favors product formation.
- Positive \( \Delta_{r} G^{\circ} \): Reaction favors reactants, not spontaneous.
Other exercises in this chapter
Problem 53
Use a mathematical equation to show how the statement leads to the conclusion cited: If a reaction is exothermic (negative \(\Delta_{1} H\) ) and if the entropy
View solution Problem 54
Use a mathematical equation to show how the statement leads to the conclusion cited: If \(\Delta_{\mathrm{r}} H\) and \(\Delta_{\mathrm{r}} S\) have the same si
View solution Problem 56
Predict whether this reaction is product-favored at \(25^{\circ} \mathrm{C}\) by calculating the change in standard Gibbs free energy from the entropy and entha
View solution Problem 73
For each reaction, an equilibrium constant at \(298 \mathrm{~K}\) is given. Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) for each reaction. $$ \begin{array}{lr}
View solution