Which of the following statements is correct for a reversible process in a state of equilibrium ?

Options

A. `DeltaG^@=2.30 RT log K`
B. `DeltaG=-2.30 RT log K`
C. `DeltaG=2.30 RT log K`
D. `DeltaG^@=-2.30 RT log K`

Answer

To solve the question regarding which statement is correct for a reversible process in a state of equilibrium, we can follow these steps:

### Step 1: Understand the Concept of Equilibrium
In a reversible process, equilibrium is reached when the rates of the forward and reverse reactions are equal. At this point, the concentrations of reactants and products remain constant over time.

**Hint:** Remember that equilibrium involves a balance between the forward and reverse reactions.

### Step 2: Recall the Gibbs Free Energy Equation
The Gibbs free energy change (ΔG) for a reaction can be expressed as:
\[ \Delta G = \Delta G^\circ + 2.30RT \log Q \]
where:
- \( \Delta G \) is the Gibbs free energy change,
- \( \Delta G^\circ \) is the standard Gibbs free energy change,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( Q \) is the reaction quotient.

**Hint:** The Gibbs free energy equation relates the spontaneity of a reaction to the concentrations of reactants and products.

### Step 3: Analyze the Condition at Equilibrium
At equilibrium, the Gibbs free energy change (ΔG) is zero, and the reaction quotient (Q) equals the equilibrium constant (K). Therefore, we can set up the equation:
\[ 0 = \Delta G^\circ + 2.30RT \log K \]

**Hint:** Equilibrium means no net change in the concentrations of reactants and products.

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ \Delta G^\circ = -2.30RT \log K \]

This equation shows the relationship between the standard Gibbs free energy change and the equilibrium constant.

**Hint:** Rearranging equations can help you find relationships between different thermodynamic quantities.

### Step 5: Identify the Correct Statement
From the derived equation, we can see that the correct statement regarding a reversible process in a state of equilibrium is:
\[ \Delta G^\circ = -2.30RT \log K \]

This matches with option D from the provided statements.

**Hint:** Look for options that express the relationship between Gibbs free energy and the equilibrium constant.

### Conclusion
The correct statement for a reversible process in a state of equilibrium is:
\[ \Delta G^\circ = -2.30RT \log K \]