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The Ksp for a very insoluble salt is 4.2×10−47 at 298 K. What is ΔG∘ for the dissolution of the salt in water?

The for a very insoluble salt is at 298 . What is for the dissolution of the salt in water?
-265 kJ/mol
-115 kJ/mol
-2.61 kJ/mol
+115 kJ/mol
+265 kJ/mol

User Enid
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2 Answers

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Final answer:

The standard Gibbs free energy change (ΔG°) for the dissolution of the very insoluble salt in water can be calculated using the solubility product constant (Ksp). At 298 K, the ΔG° is approximately -265 kJ/mol.

Step-by-step explanation:

The solubility product constant (Ksp) is a measure of the extent to which a solid solute will dissolve in water. It is related to the equilibrium constant for the dissolution reaction of the salt in water. To calculate the standard Gibbs free energy change (ΔG°) for the dissolution of the salt, we can use the equation:

ΔG° = -RTln(Ksp)

where R is the gas constant and T is the temperature in Kelvin. Plugging in the given value of Ksp = 4.2×10-47 and T = 298 K, we can calculate ΔG° to be approximately -265 kJ/mol.

User AyoO
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6 votes

Final answer:

The standard Gibbs free energy change (ΔG°) for the dissolution of a salt can be calculated using its solubility product constant (Ksp) and the formula ΔG° = -RTlnKsp, with R as the gas constant and T as the temperature in Kelvin.

Step-by-step explanation:

The question asks to calculate the standard Gibbs free energy change (ΔG°) for the dissolution of a salt in water given its solubility product constant (Ksp) at a temperature of 298 K. To find ΔG° for the dissolution, we use the formula ΔG° = -RTlnKsp, where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in kelvin (K).

First, convert the value of Ksp to a natural logarithm:

lnKsp = ln(4.2×10-47)

Then calculate the Gibbs free energy change:

ΔG° = - (8.314 J/mol·K) * (298 K) * ln(4.2×10-47)

Now solve for ΔG°, which will give you the answer in Joules per mole, and then convert it to kilojoules per mole by dividing by 1000.

The resulting value will tell us whether the dissolution process is spontaneous (negative ΔG°) or non-spontaneous (positive ΔG°).

User Jacob Turner
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