Using the thermodynamic data provided below, calculate the standard change in entropy when one mole of sodium nitrate is dissolved in water?

S° (J/K•mol)
NaNO3(s) 116.3
Na+(aq) 60.25
NO3–(aq) 146.4
Using the value you calculated, how will the solubility of sodium nitrate change if the temperature of the system is increased?

a.Solubility will increase.
b.Solubility will decrease.
c.Solubility will not be affected by temperature changes.
d.Cannot determine with provided information.

NaNO3>>Na+ + NO3-

116.3=60.25+146.4

change in Entropy= sum produts- sum reactants.

so figure the change in entropy. Notice it is postitive, so it is becoming more disordered....

the term then in the gibbs free energy equation (TdeltaS gets bigger, but it is negative)..

GibbsFreeEnergy= changeEnthalpy- TdS

so the TdS term gives a larger negative free energy, so solubility will increase. Bigger negative Gibbs, more spontaneous.

Well, let's take a look at the thermodynamic data. The standard change in entropy when one mole of sodium nitrate is dissolved in water can be calculated by subtracting the sum of the entropies of the products from the entropy of the reactant.

So, the entropy change (∆S) for the dissolution of NaNO3 can be calculated as follows:

∆S = (S°Na+(aq) + S°NO3–(aq)) - S°NaNO3(s)
= (60.25 + 146.4) - 116.3
= 90.35 J/K•mol

Now, the question is, how will the solubility of sodium nitrate change if the temperature of the system is increased?

The change in entropy (∆S) for dissolution of NaNO3 is positive, indicating that the system becomes more disordered when NaNO3 is dissolved. According to the second law of thermodynamics, an increase in temperature usually leads to an increase in entropy.

So, since increasing the temperature generally leads to an increase in entropy, we can conclude that:

a. Solubility will increase.

Good luck with your chemistry studies!

To calculate the standard change in entropy when one mole of sodium nitrate (NaNO3) is dissolved in water, we can subtract the sum of the standard molar entropies of the products from the standard molar entropy of the reactant.

Given thermodynamic data:
S°(NaNO3(s)) = 116.3 J/K•mol
S°(Na+(aq)) = 60.25 J/K•mol
S°(NO3–(aq)) = 146.4 J/K•mol

The reaction can be written as:
NaNO3(s) → Na+(aq) + NO3–(aq)

The standard change in entropy (ΔS°) can be calculated as follows:
ΔS° = ΣS°(products) - ΣS°(reactants)
= [S°(Na+(aq)) + S°(NO3–(aq))] - S°(NaNO3(s))
= [60.25 J/K•mol + 146.4 J/K•mol] - 116.3 J/K•mol
= 190.65 J/K•mol - 116.3 J/K•mol
= 74.35 J/K•mol

Therefore, the standard change in entropy when one mole of sodium nitrate is dissolved in water is 74.35 J/K•mol.

Based on this information, we can determine how the solubility of sodium nitrate will change if the temperature of the system is increased. Since the standard change in entropy is positive (74.35 J/K•mol), it indicates that the dissolution process is entropically favored. As the temperature increases, the entropy of the system also increases. According to Le Chatelier's principle, an increase in system entropy favors the dissolution process, resulting in an increase in solubility.

Therefore, the correct answer is:

a. Solubility will increase.

To calculate the standard change in entropy (∆S°) when one mole of sodium nitrate (NaNO3) is dissolved in water, we can use the following equation: ∆S° = ΣS°(products) - ΣS°(reactants).

In this case, the reactant is NaNO3(s) and the products are Na+(aq) and NO3-(aq). The given thermodynamic data provides the standard molar entropy (S°) for each species.

∆S° = [S°(Na+(aq)) + S°(NO3-(aq))] - S°(NaNO3(s))
∆S° = [60.25 J/K•mol + 146.4 J/K•mol] - 116.3 J/K•mol
∆S° = 90.35 J/K•mol

Now, let's consider how the solubility of sodium nitrate changes with temperature. The standard change in entropy (∆S°) provides information about the disorder (randomness) of the system.

If ∆S° is positive, it means that the disorder increases upon dissolution, indicating that the substance becomes more soluble at higher temperatures. In this case, ∆S° is positive, so the solubility of sodium nitrate will increase as the temperature of the system is increased.

Therefore, the correct answer is a. Solubility will increase.