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. **Understand the Relationship**: The relationship between standard 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 ideal gas constant and T is the temperature in Kelvin.

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

3. **Substitute ΔG° into the Equation**: Since ΔG° is negative, we can rewrite the equation:
\[ -2.303 RT \log K < 0 \]
This implies that the term \(-2.303 RT\) is negative (as R and T are positive constants).

4. **Determine the Sign of log K**: For the product of two terms to be negative (as in the case above), the logarithm term must be positive:
\[ \log K > 0 \]

5. **Interpret the Logarithmic Result**: If \(\log K > 0\), this means that K must be greater than 1:
\[ K > 1 \]

6. **Conclusion**: Therefore, if the standard free energy change (ΔG°) is less than zero, the equilibrium constant (K) is greater than 1, indicating that the products are favored over the reactants at equilibrium.

### Final Answer:
The equilibrium constant (K) is greater than 1.

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