Question

The standard Gibbs energy ($\Delta G^\ominus$) for the following reaction is A(s) + B$^{2+}$ (aq) $\rightleftharpoons$ A$^{2+}$ (aq) + B(s), K$_c$=10$^{12}$ at 25$^\circ$C (K$_c$= equilibrium constant)

• A. -150 kJ
• B. -96.80 kJ
• C. -68.47 kJ
• D. -100 kJ

Hint:

Relationship between standard Gibbs energy and equilibrium constant: $\Delta G^\ominus = -RT \ln K$.
If using $\log_{10}$, then $\Delta G^\ominus = -2.303 RT \log_{10} K$.
$R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.
Convert temperature to Kelvin ($T(K) = T(^\circ C) + 273.15$, or use $273$). For $25^\circ\text{C}$, $T \approx 298 \text{ K}$.
$\ln(10) \approx 2.303$.
Result will be in J/mol, convert to kJ/mol if options are in kJ.

Answer 1

The Correct Option is C

To determine the standard Gibbs energy ($\Delta G^\ominus$) for the given reaction: A(s) + B$^{2+}$ (aq) $\rightleftharpoons$ A$^{2+}$ (aq) + B(s), we use the relationship between $\Delta G^\ominus$ and the equilibrium constant ($K_c$). The equation is given by:

Δ G°=-RT\ln K_c

Where:

• Δ G° is the standard Gibbs free energy change.
• R is the universal gas constant, 8.314 J/(mol·K).
• T is the temperature in Kelvin. At 25°C, T = 298 K.
• K_c is the equilibrium constant, 10¹² in this case.

Let's substitute the values to calculate Δ G°:

Δ G° = -(8.314 J/(mol·K) × 298 K × ln(10¹²))

First, convert the logarithmic expression:

ln(10¹²) = 12 × ln(10) ≈ 12 × 2.303 = 27.636

Substitute back into the equation:

Δ G° = -8.314 × 298 × 27.636 J/mol

Δ G° ≈ -685017 J/mol = -685.017 kJ/mol

Rounding to a proper significant figure and given options, we find:

Δ G° ≈ -68.47 kJ/mol