For the reaction `H_(2)(g) + I_(2) (g) hArr 2HI (g),` the standard free energy is `DeltaG^(Theta) gt 0.` the equilibrium constant (k) would be. (A)K = 0 (B)K > 1 (C)K = 1 (D)K < 1

Options

A. `K=0`
B. `Kgt1`
C. `K=1`
D. `Klt1`

Explanation

To solve the question regarding the equilibrium constant \( K \) for the reaction

\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

given that the standard free energy change \( \Delta G^\Theta > 0 \), we can follow these steps:

### Step 1: Understand the relationship between \( \Delta G^\Theta \) and \( K \)

The standard Gibbs free energy change \( \Delta G^\Theta \) is related to the equilibrium constant \( K \) by the equation:

\[ \Delta G^\Theta = -RT \ln K \]

where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( K \) is the equilibrium constant.

### Step 2: Analyze the condition \( \Delta G^\Theta > 0 \)

Since it is given that \( \Delta G^\Theta > 0 \), we can substitute this into our equation:

\[ -RT \ln K > 0 \]

### Step 3: Rearrange the inequality

To analyze the inequality, we can multiply both sides by -1 (which reverses the inequality):

\[ RT \ln K < 0 \]

### Step 4: Solve for \( K \)

Since \( R \) and \( T \) are both positive constants, we can divide by \( RT \):

\[ \ln K < 0 \]

### Step 5: Exponentiate both sides

To remove the logarithm, we exponentiate both sides:

\[ K < e^0 \]

Since \( e^0 = 1 \):

\[ K < 1 \]

### Conclusion

Thus, the equilibrium constant \( K \) for the reaction is less than 1. Therefore, the correct answer is:

**(D) \( K < 1 \)**

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