Final temperature of acid-base reaction solution conducted in a calorimeter, inclusive of calorimeter heat capacity

Question

Two solutions, 200. mL of 0.862 M HCl(aq) and 200. mL of 0.431 M Ba(OH)₂(aq), are mixed in a constant pressure calorimeter (a coffee-cup calorimeter) that has a heat capacity of 453 J/^oC. The initial temperature of both solutions is 20.48^oC.

The heat of neutralization for this acid-base reaction is -56.2 kJ: H⁺(aq) + OH^-(aq) --> H₂O(l)

What is the final temperature of the mixed solutions?

Answer 1

In reality a constant pressure (coffee cup) calorimeter experiment differs from the idealized version in that the energy lost by the process is not totally absorbed by the water, because:

1. The coffee-cup calorimeter has a heat capacity as in J/⁰C (not a specific heat capacity as in J/g-⁰C) and it does absorb some heat. Therefore the system’s total heat is made up as follows:

q_system = q_process + q_water + q_cal

Because of 1st Law, energy of a system is always conserved. Therefore:

q_system = 0

And consequently

q_process = - (q_water + q_cal)

Now we calculate the q_process ­above using the following equations

q_water = m_water C_{p water} ΔT_water and q_cal = C_cal ΔT_cal

(Note: C_cal = heat capacity not specific heat capacity)

2. Also more realistically, q can be lost to surroundings, therefore: q_process = - (q_water + q_cal + q_surr.). However, in this problem loss to the surroundings will not be considered.

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Part 1: Determine energy transferred to final solution

Whenever a problem states quantities of two reactants, one needs to consider the possibility of a limiting reagent. In this case a quick mental calculation indicates that the reactant molar amounts are in the proper proportions, as per the balanced equation. The mole ratio between HCl and Ba(OH)₂ is 2 to 1.

2HCl(aq) + Ba(OH)₂(aq) --> BaCl₂(aq) + 2H₂O(l)

Hence the amount of energy transferred (q) to the final solution is determined by basing the calculation on the amount of either reactant, I chose to base the calculation on Ba(OH)₂ to avoid having to divide the heat of neutralization by 2 by basing it on HCl.

q_{absorbed by solution} = +56,2 kJ (0.200 L sln)(0.431 mol Ba(OH)₂ / 1 L sln) = +4.84 kJ

Part 2: Determine the final temperature of the mixed solution



q_{acid-base reaction} = -(q_solution + q_{cal) =} [mass_solution (C_solution ) (T_final - T_initial)] + [q _cal (T_final - T_initial)]

-4,840 J = -[400. g (4.18 J/g^oC) (T_final - 20.48^oC) + 438 J/^oC (T_final - 20.48^oC)]

+4,840 J = 1,672 J/^oC (T_final - 20.48^oC) + 438 J/^oC (T_final - 20.48^oC)

+4,840 J = 2,110 J/^oC (T_final - 20.48^oC)

(T_final - 20.48^oC) = 2.29^oC

T_final = 20.48^oC + 2.29^oC= 22.77^oC