Based on the thermodynamic data, which of the following is true at 298 K? — Thermodynamics Chemistry Question

STEPS:

1. Identify the relevant thermodynamic variable from the table: On page 25 of the 2017 exam booklet, the thermodynamic data table for this peptide-coupling reaction provides the standard Gibbs free energy change at $298\text{ K}$:
$$\Delta G^\circ_{298} = \mathbf{+15\text{ kJ/mol}_{\text{rxn}}}$$

2. Recall the mathematical relationship between standard free energy change and the equilibrium constant:
The equation that relates standard Gibbs free energy change ($\Delta G^\circ$) to the thermodynamic equilibrium constant ($K_{\text{eq}}$) is:
$$\Delta G^\circ = -RT \ln K_{\text{eq}}$$
where $R$ is the ideal gas constant ($8.314\text{ J/(mol K)}$), $T$ is the absolute temperature in Kelvin ($298\text{ K}$), and $\ln$ is the natural logarithm.

3. Analyze the sign of the terms mathematically:
* Rearranging the formula to solve for the natural log of the equilibrium constant gives:
$$\ln K_{\text{eq}} = \frac{-\Delta G^\circ}{RT}$$
* Because temperature ($T = 298\text{ K}$) and the gas constant ($R$) are always positive, and our $\Delta G^\circ$ is positive ($+15\text{ kJ/mol}$), the quotient on the right must be negative:
$$\ln K_{\text{eq}} = \frac{-(+15\text{ kJ/mol})}{RT} < 0$$

4. Solve for the range of the equilibrium constant ($K_{\text{eq}}$):
* To isolate $K_{\text{eq}}$, take the exponential of both sides:
$$K_{\text{eq}} = e^{\text{negative number}}$$
* Mathematically, the constant $e$ ($\approx 2.718$) raised to any negative exponent ($e^{-x} = \frac{1}{e^x}$) always yields a value strictly between 0 and 1:
$$\mathbf{0 < K_{\text{eq}} < 1}$$

5. Conclude:
This shows that at $298\text{ K}$, the equilibrium constant is greater than 0 but less than 1, confirming Option B as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect ($K_{\text{eq}} = 0$): An equilibrium constant can never equal exactly 0. A value of $K_{\text{eq}} = 0$ would mathematically require an infinite positive free energy ($\Delta G^\circ \rightarrow +\infty$) and would physically represent a reaction that is completely unable to form any product whatsoever under any circumstances, which is not possible in dynamic chemical systems.
• Option C is incorrect ($K_{\text{eq}} = 1$): For $K_{\text{eq}}$ to equal exactly 1, the natural log of $K_{\text{eq}}$ would be 0 ($\ln(1) = 0$). According to the relationship $\Delta G^\circ = -RT \ln K_{\text{eq}}$, this would require a standard Gibbs free energy change of exactly $0\text{ kJ/mol}$, which is not the case for this reaction.
• Option D is incorrect ($K_{\text{eq}} > 1$): For $K_{\text{eq}}$ to be greater than 1, its natural log must be positive ($\ln K_{\text{eq}} > 0$). This occurs only when the standard Gibbs free energy change is negative ($\Delta G^\circ < 0$), representing a thermodynamically favored (spontaneous) reaction. Because this reaction is thermodynamically unfavored ($\Delta G^\circ = +15\text{ kJ/mol}$), reactants are favored at equilibrium, and $K_{\text{eq}}$ must be less than 1.