Reaction that have standard free energy changes less than zero always have equilibrium constant equal to

Options

A. unity
B. greater than unity
C. less than unity
D. zero

Answer

To solve the question regarding the relationship between standard free energy change (ΔG°) and the equilibrium constant (K), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relationship between standard Gibbs free energy change (ΔG°) and the equilibrium constant (K) is given by the equation:
\[ \Delta G° = -2.303 RT \log K \]
where R is the universal gas constant and T is the temperature in Kelvin.

2. **Analyzing the Given Condition**:
The question states that ΔG° is less than zero (ΔG° < 0). This indicates that the reaction is spontaneous under standard conditions.

3. **Substituting the Condition into the Equation**:
Since ΔG° is negative, we can rewrite the equation:
\[ -2.303 RT \log K < 0 \]
Here, both R (the gas constant) and T (temperature) are positive values. Therefore, the product \( -2.303 RT \) is negative.

4. **Determining the Sign of log K**:
For the entire left side of the inequality to be negative, the term \( \log K \) must be positive. This is because a negative number multiplied by a positive number (like \( -2.303 RT \)) will yield a negative result only if \( \log K \) is positive.

5. **Concluding the Value of K**:
If \( \log K > 0 \), then K must be greater than 1 (since \( \log 1 = 0 \)). Therefore, we conclude that:
\[ K > 1 \]

6. **Final Answer**:
Thus, when the standard free energy change (ΔG°) is less than zero, the equilibrium constant (K) is greater than 1.

### Conclusion:
The correct answer to the question is that the equilibrium constant (K) is greater than 1 when ΔG° is less than 0.

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