Which of the following graph correctly represent for equilibrium reaction whose `K_(p) gt 1`?

Options

A.
B.
C.
D.

Answer

To determine which graph correctly represents an equilibrium reaction where \( K_p > 1 \), we can follow these steps:

### Step 1: Understanding \( K_p \)
- The equilibrium constant \( K_p \) is a measure of the ratio of the concentrations of products to reactants at equilibrium. When \( K_p > 1 \), it indicates that at equilibrium, the concentration of products is greater than that of reactants.

### Step 2: Relating \( K_p \) to Gibbs Free Energy
- The relationship between Gibbs free energy (\( \Delta G \)) and \( K_p \) is given by the equation:
\[ \Delta G = -RT \ln K_p \]
where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. If \( K_p > 1 \), then \( \ln K_p \) is positive, which makes \( \Delta G \) negative.

### Step 3: Implications of \( \Delta G < 0 \)
- A negative \( \Delta G \) indicates that the reaction is spontaneous in the forward direction. This means that the products are favored over the reactants at equilibrium.

### Step 4: Analyzing the Graphs
- We need to analyze the graphs provided to see which one reflects the relationship where the Gibbs free energy of the products is lower than that of the reactants.

1. **Graph A**: Shows that the Gibbs free energy of products is lower than that of reactants. This is consistent with \( K_p > 1 \).
2. **Graph B**: Shows that the Gibbs free energy of reactants is lower than that of products, which contradicts our condition.
3. **Graph C**: Indicates that the Gibbs free energy of reactants and products are equal, which is not the case for \( K_p > 1 \).
4. **Graph D**: Shows that the Gibbs free energy of reactants is higher than that of products, which is also consistent with \( K_p > 1 \).

### Step 5: Conclusion
- Between Graph A and Graph D, Graph A correctly shows the trend where the Gibbs free energy of products is lower than that of reactants, indicating a spontaneous reaction favoring products. Therefore, the correct answer is **Graph A**.