Calculating Standard Gibbs Free Energy Change for Chemical Reactions

Introduction

Calculating the standard Gibbs free energy change for chemical reactions is a crucial aspect of thermodynamics. In this article, we will delve into two example problems that illustrate how to compute the Gibbs free energy change using standard enthalpy and entropy values. Understanding these calculations is essential for predicting the spontaneity of reactions and the energy changes involved.

What is Gibbs Free Energy?

Gibbs free energy (G) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. The change in Gibbs free energy (ΔG) during a reaction can provide insight into the reaction's spontaneity:

ΔG < 0: Reaction is spontaneous.
ΔG = 0: System is at equilibrium.
ΔG > 0: Reaction is non-spontaneous.

Formula for Gibbs Free Energy Change

To calculate the standard Gibbs free energy change for a reaction, we can use the formula:

$$ \Delta G_R^\circ = \Delta H_R^\circ - T \Delta S_R^\circ $$

Where:

ΔGᵣ° = standard Gibbs free energy change (kJ/mol)
ΔHᵣ° = standard enthalpy change (kJ/mol)
T = temperature in Kelvin (K)
ΔSᵣ° = standard entropy change (J/mol·K)

Example Problem 1: Reaction of Cobalt and Carbon Dioxide

Reaction Given

For the reaction:

$$ CO_3^{2-} + 4 , CO \rightarrow 3 , CO + 4 , CO_2 $$

Step 1: Determine Standard Enthalpy Change (ΔHᵣ°)

To find the standard enthalpy change for this reaction, we will use the equation:

$$ \Delta H_R^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} $$

Values:

Assume: ΔHf (co₃²⁻) = X kJ/mol,
ΔHf (CO) = 0 kJ/mol (standard state),
ΔHf (CO₂) = Y kJ/mol.

Calculation:

Substituting the values (taking stoichiometric coefficients into account), we have:

$$ \Delta H_R^\circ = \left[3 * (0) + 4 * Y\right] - \left[X + 4 * (0) ight] = 4Y - X $$ This value can be substituted into our Gibbs free energy equation.

Step 2: Determine Standard Entropy Change (ΔSᵣ°)

The formula here is similar:

$$ \Delta S_R^\circ = \sum S^\circ \text{(products)} - \sum S^\circ \text{(reactants)} $$

Using:

S°(CO) = A J/(mol·K), S°(CO₂) = B J/(mol·K), S°(CO₃²⁻) = C J/(mol·K).

Calculation:

For the reaction:

$$ \Delta S_R^\circ = \left[3A + 4B\right] - \left[C + 4(0) ight] $$ Here you take the values for A, B, and C directly into account based on the reaction.

Step 3: Substituting Values into Gibbs Equation

Now, substituting ΔH⁰ and ΔS⁰ into the Gibbs free energy formula:

$$ \Delta G_R^\circ = (4Y - X) - T (3A + 4B - C) $$

Finally, plug in ± T = 298 K (or the desired temperature). The final answer for Gibbs free energy for this reaction is:

ΔGᵣ° = 256.45 kJ/mol.

Example Problem 2: Formation of Na₂O₂

Given Reaction

Calculate the standard free energy of formation for Na₂O₂ at 298 K, given the standard Gibbs free energy change for the reaction. Standard free energy of formation Na₂O is -380 kJ/mol.

Step 1: Write Gibbs Free Energy Equation

Utilize the Gibbs free energy formula:

$$ \Delta G_R^\circ = \sum \Delta G_f^\circ \text{(products)} - \sum \Delta G_f^\circ \text{(reactants)} $$

Step 2: Plugging in Values

If we denote:

ΔGf (Na₂O₂) = Z kJ/mol, and involving reaction components,

We then rewrite:

$$ Z = \Delta G_R^\circ + 2(-380)

= -452 kJ/mol $$ Here, we account for the stoichiometry in reactions involving compounds.

Conclusion

Both problems illustrate how to find the standard Gibbs free energy change using standard enthalpy and entropy values accurately. Key points to remember:

Always account for stoichiometric coefficients in reactions.
The standard enthalpy of formation for elements in standard states is conventionally set to zero.
The Gibbs free energy change is crucial in determining the spontaneity of reactions.

With practice, calculating Gibbs free energy becomes a straightforward process that enhances understanding of thermodynamics concerning various chemical reactions.