Calculate ΔG at 25 ºC for a reaction in which Ca2+(aq) combines with CO32–(aq) to form a precipitate of CaCO3(s) if the concentrations of Ca2+(aq) and CO32–(aq) are 0.023 M and 0.13 M, respectively. (ΔGº = –47.94 kJ)
A. –62.3 kJ/mol
B. –33.5 kJ/mol
C. –17.8 kJ/mol
D. –46.7 kJ/mol
E. –47.8 kJ/mol
DG = -47940 + RT*ln[1/(0.023*0.13)]
To calculate ΔG at 25 ºC, we can use the equation:
ΔG = ΔGº + RT * ln (Q)
Where:
- ΔG is the change in Gibbs free energy.
- ΔGº is the standard Gibbs free energy change.
- R is the gas constant (8.314 J/mol·K).
- T is the temperature in Kelvin.
- Q is the reaction quotient.
First, we need to calculate the reaction quotient, Q. This is determined by the concentrations of the reactants and products at a given point in the reaction.
The balanced equation for the reaction is:
Ca2+(aq) + CO32–(aq) → CaCO3(s)
Since the reaction results in the formation of a precipitate of CaCO3(s), the concentrations of the solid products are not included in the calculation of Q.
Thus, Q = [Ca2+] * [CO32–]
Plugging in the given concentrations:
Q = (0.023 M) * (0.13 M) = 0.00299 M^2
Next, we can plug the values into the equation for ΔG:
ΔG = -47.94 kJ/mol + (8.314 J/mol·K) * (298 K) * ln (0.00299)
Calculating the natural logarithm:
ln (0.00299) ≈ -5.81
Substituting this value back into the equation:
ΔG ≈ -47.94 kJ/mol + (8.314 J/mol·K) * (298 K) * (-5.81)
ΔG ≈ -47.94 kJ/mol + (-14479 J/mol)
Converting J to kJ:
ΔG ≈ -47.94 kJ/mol - 14.48 kJ/mol
ΔG ≈ -62.42 kJ/mol
Rounding to two decimal places, the value of ΔG is approximately -62.42 kJ/mol.
Therefore, the correct answer is option A: -62.3 kJ/mol.