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Question

A 200. mL of 0.862 M  HCl is mixed with 200. mL of 0.431 M  Ba(OH)2 in a constant pressure coffee cup calorimeter. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 24.48°C. What is the heat of reaction (enthalpy) per mole of acid? Assume the mixture has a specific heat of 4.18 J/g·°C and that the densities of the reactant solutions are both 1.00 g/mL. calculate the final temperature of the mixed solution. The heat of neutralization is -56.2kJ/mol. ​

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Answer 1

2HCl + Ba(OH)2 ==> 2H2O + BaCl2 ... BALANCED EQUATION

q = mC∆T

q = heat = -56.2 kJ/mole x moles (see below)

m = mass = 400 ml x 1 g/ml = 400 g

C = specific heat = 4.18 J/gº

∆T = change in temperature = ?

To find q for the reaction, let's see how many moles of reactants we have...

moles HCl = 0.200 L x 0.862 mol/L = 0.1724 mol HCl

moles Ba(OH)2 = 0.200 L x 0.431 mol/L = 0.0862

Complete neutralization because the mol ratio of HCl : Ba(OH)2 is 2:1 as is the number of moles.

∆H = -56.2 kJ/mol x 0.0862 moles = -4.84 kJ = 4840 J

q = mC∆T

4840 J = (400 g)(4.18 J/gº)(∆T)

∆T = 2.89º

Since ∆H is negative, this indicates an exothermic reaction so we must ADD 2.89º to the original temperature.

Final temperature = 24.48º + 2.89º = 27.37ºC