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What is the ΔG (kJ/mol) for a reaction at 25 Celsius that is:

Mg3(PO4)2 (s) ⇄ 3 Mg2+ (aq) + 2 PO43− (aq) ΔG0 = 137.0 kJ/mol
If there is initially 0.65 M Mg2+(aq) and 0.43 M PO43− (aq) in solution?

1 Answer

4 votes

Final answer:

The ΔG (kJ/mol) for the given reaction at 25 Celsius can be calculated using the equation ΔG = ΔG° + RTln(Q), where Q is the reaction quotient calculated using the initial concentrations of Mg2+ and PO43−. Given the ΔG° value of 137.0 kJ/mol, the value of ΔG for the reaction is -97.2 kJ/mol.

Step-by-step explanation:

The ΔG (kJ/mol) for a reaction at 25 Celsius can be calculated using the equation:



ΔG = ΔG° + RTln(Q)



where ΔG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.



In this case, the given reaction is:



Mg3(PO4)2 (s) ⇄ 3 Mg2+ (aq) + 2 PO43− (aq)



To calculate ΔG, we need to know the value of Q, which can be calculated using the initial concentrations of Mg2+ and PO43− in solution.



Q = [Mg2+]3[PO43−]2



Q = (0.65 M)3(0.43 M)2 = 0.011 M5



Given that ΔG° = 137.0 kJ/mol, R = 8.314 J/(mol⋅K), and T = 298 K, we can calculate the value of ΔG.



ΔG = 137.0 kJ/mol + (8.314 J/(mol⋅K) × 298 K × ln(0.011 M5)) = -97.2 kJ/mol



Therefore, the value of ΔG for the reaction at 25 Celsius is -97.2 kJ/mol.

answered
User Chase Ries
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