A calorimeter, with calorimeter constant 7.31 J K–1, was used to measure the heat of reaction for mixing 100.0 mL of 0.100 M Ag+ with 100.0 mL of 0.100 M Cl–. The temperature change was +0.770 K for the 200.0 g of solution (specific heat capacity = 4.35 J K–1g–1, density = 1.00 g cm– 3) and the calorimeter.
(i) What is the reaction occurring in the calorimeter? (ii) How much heat was absorbed by the calorimeter? (iii)How much heat was absorbed by the solution? (iv) How much heat was released by the reaction?
(v) What is the heat of reaction per mole of product formed?
a. Ag^+(aq) + Cl^-(aq) ==> AgCl(s)
b. q absorbed by calorimeter = Ccal*dT = 7.31 J/K x 0.770 K = ?
c. q absorbed by soln = mcdT = 200 x 4.35 x 0.770 = ?
d. dHrxn = b + c
e. Calculate mols AgCl formed in the reaction and dH in part d will be heat for that many mols AgCl formed. Convert that to dH/1 mol.
Post your work if you get stuck.
Can you please show the working out for e?
To answer these questions, we need to understand the concepts of heat, specific heat capacity, density, and calorimetry.
(i) What is the reaction occurring in the calorimeter?
The reaction occurring in the calorimeter is the mixing of Ag+ and Cl- ions. This forms AgCl, which is a white precipitate.
(ii) How much heat was absorbed by the calorimeter?
We can calculate the heat absorbed by the calorimeter using the equation:
q_calorimeter = C_calorimeter * ΔT
where q_calorimeter is the heat absorbed by the calorimeter, C_calorimeter is the calorimeter constant, and ΔT is the change in temperature. Plugging in the values:
q_calorimeter = 7.31 J K^(-1) * 0.77 K = 5.63 J
So, the calorimeter absorbed 5.63 J of heat.
(iii) How much heat was absorbed by the solution?
To calculate the heat absorbed by the solution, we need to consider both the heat absorbed by the solution itself and the heat absorbed due to the temperature change.
First, we calculate the heat absorbed by the solution itself:
q_solution = mass_solution * specific heat capacity_solution * ΔT
Given that the mass of the solution is 200.0 g (100.0 mL has the same mass as 100.0 g due to the density of 1.00 g cm^(-3)), the specific heat capacity is 4.35 J K^(-1) g^(-1), and ΔT is 0.770 K, we can calculate:
q_solution = 200.0 g * 4.35 J K^(-1) g^(-1) * 0.770 K = 671.10 J
Therefore, the heat absorbed by the solution is 671.10 J.
(iv) How much heat was released by the reaction?
To determine the heat released by the reaction, we need to assume that the heat absorbed by the solution is equal to the heat released by the reaction (assuming no heat is lost to the surroundings).
q_reaction = q_solution = 671.10 J
Therefore, the heat released by the reaction is 671.10 J.
(v) What is the heat of reaction per mole of product formed?
To calculate the heat of reaction per mole of product formed, we need to determine the number of moles of product formed in the reaction. In this case, AgCl is formed.
From the balanced chemical equation for the reaction:
Ag+ (aq) + Cl- (aq) → AgCl (s)
We can see that the stoichiometry is 1:1, meaning 1 mole of AgCl is formed for every 1 mole of Ag+ and Cl- ions reacted.
Therefore, the moles of AgCl formed in the reaction is the same as the moles of Ag+ and Cl- ions reacted.
To calculate the moles of ions reacted, we can use the equation:
moles = concentration * volume
Given that the volumes used were both 100.0 mL and the concentrations were both 0.100 M, the moles of ions reacted are both:
moles = 0.100 M * 0.100 L = 0.0100 moles
Since the stoichiometry is 1:1, we can conclude that 0.0100 moles of AgCl was formed.
Finally, we can calculate the heat of reaction per mole of product formed using the equation:
ΔH = q_reaction / moles
Plugging in the values:
ΔH = 671.10 J / 0.0100 moles = 67,110 J/mol
So, the heat of reaction per mole of AgCl formed is 67,110 J/mol.