In an equilibrium reaction for which `DeltaG=0,` the equlibrium constant K =

Options

A. 0
B. 1
C. 2
D. 10

Explanation

To find the equilibrium constant \( K \) for a reaction where \( \Delta G = 0 \), we can follow these steps:

### Step 1: Understand the relationship between \( \Delta G \) and \( K \)
The Gibbs free energy change (\( \Delta G \)) for a reaction is related to the equilibrium constant (\( K \)) by the equation:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( Q \) is the reaction quotient.

### Step 2: Set \( \Delta G \) to 0
At equilibrium, \( \Delta G = 0 \). Therefore, we can rewrite the equation as:
\[ 0 = \Delta G^\circ + RT \ln Q \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ \Delta G^\circ = -RT \ln Q \]

### Step 4: Relate \( Q \) to \( K \)
At equilibrium, the reaction quotient \( Q \) is equal to the equilibrium constant \( K \). Thus, we can write:
\[ \Delta G^\circ = -RT \ln K \]

### Step 5: Analyze the case when \( \Delta G^\circ = 0 \)
If \( \Delta G^\circ = 0 \), then we have:
\[ 0 = -RT \ln K \]
This implies that:
\[ RT \ln K = 0 \]

### Step 6: Solve for \( K \)
Since \( RT \) is not zero (as long as the temperature is above absolute zero), we can conclude that:
\[ \ln K = 0 \]
Exponentiating both sides gives us:
\[ K = e^0 = 1 \]

### Final Answer
Thus, the equilibrium constant \( K \) when \( \Delta G = 0 \) is:
\[ K = 1 \]

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