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The value of DeltaG^(@) for a reaction i...

The value of `DeltaG^(@)` for a reaction in aqueous phase having `K_(c)=1,` would be :

A

`-RT`

B

`-1`

C

`0`

D

`+RT`

Text Solution

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The correct Answer is:
To find the value of ΔG° for a reaction in the aqueous phase with Kc = 1, we can use the following steps: ### Step-by-Step Solution: 1. **Understand the Relationship**: The relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (Kc) is given by the equation: \[ \Delta G^\circ = -2.303 \log_{10}(K_c) \] 2. **Substitute the Given Value**: Since the problem states that Kc = 1, we substitute this value into the equation: \[ \Delta G^\circ = -2.303 \log_{10}(1) \] 3. **Calculate the Logarithm**: We know that the logarithm of 1 is 0: \[ \log_{10}(1) = 0 \] 4. **Substitute the Logarithm Result**: Now, substitute the value of the logarithm back into the equation: \[ \Delta G^\circ = -2.303 \times 0 \] 5. **Simplify the Expression**: Multiplying by zero gives: \[ \Delta G^\circ = 0 \] 6. **Conclusion**: Therefore, the value of ΔG° for the reaction is: \[ \Delta G^\circ = 0 \] ### Final Answer: The value of ΔG° for the reaction is **0**. ---
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