The cell in which the following reaction occurs has at 298 K. The standard Gibbs energy will be

To determine the standard Gibbs energy change ($\Delta G^\circ$) for a reaction occurring in a cell at 298 K, we generally use the following equation derived from the relationship of Gibbs free energy, enthalpy, and entropy:

$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$

Where:

• $\Delta G^\circ$ is the standard Gibbs energy change,
• $\Delta H^\circ$ is the standard enthalpy change,
• $T$ is the temperature in Kelvin (298 K in this case),
• $\Delta S^\circ$ is the standard entropy change.

1. Break Down the Problem

We need the values for $\Delta H^\circ$ and $\Delta S^\circ$ for the specific reaction you are investigating in order to compute $\Delta G^\circ$.

2. Relevant Concepts

• The reaction's enthalpy change ($\Delta H^\circ$) can often be found in thermodynamic tables.
• The entropy change ($\Delta S^\circ$) can also be found in tables or calculated from standard entropy values of reactants and products.

3. Analysis and Detail

For example, if we have the standard enthalpy change ($\Delta H^\circ$) as -100 kJ/mol and the standard entropy change ($\Delta S^\circ$) as -200 J/mol·K, we will first convert $\Delta S^\circ$ into kJ for coherence in units:

$\Delta S^\circ = -200 \, \text{J/mol·K} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -0.2 \, \text{kJ/mol·K}$

Now substituting the values into the Gibbs energy equation:

$\Delta G^\circ = (-100 \, \text{kJ/mol}) - (298 \, \text{K})(-0.2 \, \text{kJ/mol·K})$

Calculating the temperature and entropy component:

$\Delta G^\circ = -100 \, \text{kJ/mol} + 59.6 \, \text{kJ/mol} = -40.4 \, \text{kJ/mol}$

4. Verify and Summarize

We have calculated $\Delta G^\circ$ using the hypothetical values provided. It is essential to substitute the actual thermodynamic values of the reaction you are inquiring about to perform this calculation accurately.

Final Answer

The standard Gibbs energy change for the reaction at 298 K can be computed using the provided values. If $\Delta H^\circ = -100 \, \text{kJ/mol}$ and $\Delta S^\circ = -200 \, \text{J/mol·K}$, then:

$\Delta G^\circ = -40.4 \, \text{kJ/mol}$

Please provide the specific values for the enthalpy and entropy changes for the exact computation.