For a reaction that has an equilibrium constant of 9 × 10–9, which of the following statements must be true?
A) ∆G° is positive.
B) ∆G° is negative.
C) ∆H° is negative.
D) ∆H° is positive.
E) ∆S° is positive.
To determine the correct statement, we need to understand the relationship between the equilibrium constant (K) and the standard Gibbs free energy change (∆G°), enthalpy change (∆H°), and entropy change (∆S°).
The equation that relates these quantities is as follows:
∆G° = -RT ln(K)
where ∆G° is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and ln represents the natural logarithm.
Now let's analyze the answer choices:
A) ∆G° is positive.
If the equilibrium constant is less than 1 (as in this scenario, where K = 9 × 10–9), the natural logarithm of K will be negative. Multiplying a negative value by a positive constant (–RT) will result in a positive ∆G°. Therefore, option A is incorrect.
B) ∆G° is negative.
As explained above, if the equilibrium constant is less than 1, ∆G° will be positive. Thus, option B is incorrect.
C) ∆H° is negative.
The value of the equilibrium constant itself (∆G° = -RT ln(K)) does not provide information about the sign of ∆H°. The answer choice does not make a direct connection between K and ∆H°, so we cannot determine its sign directly from the equilibrium constant. Therefore, option C is incorrect.
D) ∆H° is positive.
Similar to option C, the equilibrium constant does not directly indicate the sign of ∆H°. Therefore, option D is incorrect.
E) ∆S° is positive.
The equation relating ∆G° and K (∆G° = -RT ln(K)) tells us that if K is less than 1, ∆G° will be positive. This, in turn, implies that the natural logarithm of K is negative. To obtain a negative logarithm, K must be less than 1, which implies that ∆S° is positive. Therefore, option E is correct.
In conclusion, the correct statement is E) ∆S° is positive.
dGo = -RTlnK
Solve it. See if dGo is + or -