25.0 mL of a 0.50 M solution of acid HA was combined with 25.0 mL of a 0.50 solu
ID: 1057544 • Letter: 2
Question
25.0 mL of a 0.50 M solution of acid HA was combined with 25.0 mL of a 0.50 solution of base MOH in a calorimeter with a calorimeter constant of 13.5 J/C. The initial temperature of the solution was 23.3 C and the maximum temperature was 34.7 C. The resulting solution had a specific heat capacity of 3.92 J/g·C and a density of 1.04 g/mL. Calculate each of the following
a) The mass of the resulting solution.
b) The heat absorbed by the solution.
c) The heat absorbed by the calorimeter
d) The heat released by the reaction
e) The enthalpy change of the reaction
Explanation / Answer
Q = m c Delta T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g oC)
Delta T = change in temperature = Tfinal - Tinitial in oC
The mass of the resulting solution. = 25 ml + 25 ml = 50 ml x 1.04 = 52 gm
Heat produced by reaction = Heat absorbed by solution + Heat absorbed by Calorimeter
) The heat absorbed by the solution.
Q = 52 x 3.92 x (34.7 - 23.3 )
Q = 2323.8 Joules
The heat absorbed by the calorimeter
Q = (34.7 - 23.3 ) x 13.5 = 153.9 Joules
The heat released by the reaction
Q = 2323.8 + 153.9 = 2477.7 Joules
SInce it is exothermic reaction We can use negative before energy
Q = -2477.7 Joules
The enthalpy change of the reaction
Moles of salt formed = 25 x 0.05 /1000 = 0.00125 Moles
enthalpy change = -2477.7 /0.00125 = 1982160 Joules / Mole or 1982.2 Kj / Moles
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