For the reaction, `H_2(g) + I_2(g) hArr 2HI(g)` , the standard free energy is `DeltaG^(0) gt 0`. The equilibrium constant (K) would be

Options

A. `K=0`
B. `K gt 1`
C. `K=1`
D. `K lt 1`

Explanation

To determine the equilibrium constant (K) for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) given that the standard free energy change (\( \Delta G^{\circ} \)) is greater than 0, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between \( \Delta G^{\circ} \) and K**:
The relationship between the standard free energy change (\( \Delta G^{\circ} \)) and the equilibrium constant (K) is given by the equation:
\[ \Delta G^{\circ} = -RT \ln K \]
where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

2. **Analyze the Given Information**:
We know that \( \Delta G^{\circ} > 0 \). This indicates that the reaction is non-spontaneous under standard conditions.

3. **Interpret the Equation**:
Since \( \Delta G^{\circ} \) is positive, we can infer the following from the equation:
\[ -RT \ln K > 0 \]
This implies that \( \ln K < 0 \) because \( -RT \) is always negative (as R and T are positive).

4. **Determine the Value of K**:
If \( \ln K < 0 \), it follows that:
\[ K < 1 \]
This means that at equilibrium, the concentration of the reactants is greater than the concentration of the products.

5. **Conclusion**:
Therefore, since \( K < 1 \), the correct answer to the question is that the equilibrium constant (K) would be less than 1.

### Final Answer:
The equilibrium constant (K) would be less than 1.

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