A reaction is peformed in which 50 ml of .400M silver nitrate and 75 ml of .100M hydrochloric acid are mixed together. The initial temperature of the solutionn was 25 C and the final temperature was 27.2 C. What is the enthalpy, in kj/mol, for the formation of the precipitate (silver chloride)? The volume of the final solution was 125 ml, and it had the density of 1.00g/ml. The specific heat of solution is 4.18 J/g*C.

Question

A reaction is peformed in which 50 ml of .400M silver nitrate and 75 ml of .100M hydrochloric acid are mixed together. The initial temperature of the solutionn was 25 C and the final temperature was 27.2 C. What is the enthalpy, in kj/mol, for the formation of the precipitate (silver chloride)? The volume of the final solution was 125 ml, and it had the density of 1.00g/ml. The specific heat of solution is 4.18 J/g*C.

Answer 1

q = heat generated = [mass H2O x specific heat H2O x (Tfinal-Tinitial)]

That gives you delta H for the reaction.
q/mol AgCl = delta H/mol
Note: It appears HCl is the limiting reagent.

Answer 2

To determine the enthalpy change (ΔH) for the formation of the precipitate (silver chloride), we can use the formula:

ΔH = mcΔT / moles

where:
m = mass of the solution in grams
c = specific heat of the solution in J/g°C
ΔT = change in temperature in °C
moles = moles of AgCl formed

To find the moles of AgCl formed, we need to use the stoichiometry of the balanced chemical equation between AgNO3 and HCl, assuming the reaction goes to completion:

AgNO3 + HCl -> AgCl + HNO3

The stoichiometric ratio between AgNO3 and AgCl is 1:1, which means that 1 mole of AgNO3 reacts to produce 1 mole of AgCl.

First, let's calculate the moles of AgNO3:

moles of AgNO3 = volume (L) × concentration (M)
= 0.050 L × 0.400 mol/L
= 0.020 mol

Similarly, let's calculate the moles of HCl:

moles of HCl = volume (L) × concentration (M)
= 0.075 L × 0.100 mol/L
= 0.0075 mol

Since AgNO3 and HCl react in a 1:1 ratio, the limiting reactant is HCl, and therefore, 0.0075 mol of AgCl is formed.

Now, we need to calculate the mass of the solution:

mass = volume × density
= 0.125 L × 1.00 g/mL
= 125 g

Next, we can calculate the heat change (q) using the formula:

q = mcΔT

q = (125 g) × (4.18 J/g°C) × (27.2°C - 25°C)
≈ 1076 J

Finally, we can calculate the enthalpy change (ΔH) in kJ/mol:

ΔH = q / moles
= 1076 J / 0.0075 mol
≈ 143,466.67 J/mol
≈ 143.47 kJ/mol (rounded to two decimal places)

Therefore, the enthalpy change for the formation of the precipitate (silver chloride) is approximately 143.47 kJ/mol.