The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, and ΔG for the boiling process at this temperature

The question is incomplete; the complete question is;

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the

signs of ∆H, ∆S, and ∆G for the boiling process at this temperature.

A. ∆H > 0, ∆S > 0, ∆G < 0

B. ∆H > 0, ∆S > 0, ∆G > 0

C. ∆H > 0, ∆S < 0, ∆G < 0

D. ∆H < 0, ∆S > 0, ∆G > 0

E. ∆H < 0, ∆S < 0, ∆G > 0

∆H > 0, ∆S > 0, ∆G < 0

Explanation:

If we look at the question carefully, we will observe that it deals with a phase change from liquid to vapour phase.

Energy is required to break the intermolecular bonds in the liquid as it changes into vapour hence the process is endothermic, ∆H>0.

Also, the entropy of the vapour phase is greater than that of the liquid phase hence there is a positive change in entropy, ∆S>0.

Lastly, the process is spontaneous, hence the change in free energy ∆G is less than zero.

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%