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A large positive value of DeltaG^(ɵ) cor...

A large positive value of `DeltaG^(ɵ)` corresponds to which of these?

A

Small positive `K`

B

Small negative `K`

C

Large positive `K`

D

Large negative `K`

Text Solution

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The correct Answer is:
To determine the significance of a large positive value of ΔG° (Gibbs free energy change at standard conditions), we can analyze the relationship between ΔG° and the equilibrium constant (K_eq). ### Step-by-Step Solution: 1. **Understand the Relationship**: The relationship between ΔG° and K_eq is given by the equation: \[ \Delta G^\circ = -RT \ln K_{eq} \] where R is the universal gas constant and T is the temperature in Kelvin. 2. **Analyze the Equation**: - If K_eq is greater than 1, then ln K_eq is positive, making ΔG° negative. This indicates that the reaction favors the formation of products. - If K_eq is equal to 1, then ln K_eq is zero, and ΔG° is zero, indicating that the system is at equilibrium. - If K_eq is less than 1, then ln K_eq is negative, making ΔG° positive. This indicates that the reaction favors the formation of reactants. 3. **Large Positive ΔG°**: A large positive value of ΔG° indicates that the reaction strongly favors the reactants over the products. This corresponds to a very small value of K_eq. 4. **Conclusion**: Therefore, a large positive value of ΔG° corresponds to a small positive value of K_eq. ### Final Answer: A large positive value of ΔG° corresponds to a small positive value of K_eq.

To determine the significance of a large positive value of ΔG° (Gibbs free energy change at standard conditions), we can analyze the relationship between ΔG° and the equilibrium constant (K_eq). ### Step-by-Step Solution: 1. **Understand the Relationship**: The relationship between ΔG° and K_eq is given by the equation: \[ \Delta G^\circ = -RT \ln K_{eq} \] ...
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