ICIC 016: Enthalpy of Neutralizatiorn 19 NAME SECTIONDATE Pre-Laboratory Assignm
ID: 575894 • Letter: I
Question
ICIC 016: Enthalpy of Neutralizatiorn 19 NAME SECTIONDATE Pre-Laboratory Assignment 1) When a 100.0 mL sample of hot water at 45.0°C was mixed with a 100.0 mL sample of cold water at 19.8°C in a coffee cup calorimeter, the final (maximum) temperature of the mixture was 29.7 Using Equations 9-13, calculate the calorimeter constant for this calorimeter. Show all work. I/g.°C and its density was 1.03 g/mL. The calorimeter constant was 3.25 J/°C. a) Calculate the Hneutraluation for the reaction of NH3 and acetic acid. Show all work 2) When a neutralization reaction was carried out using 100.0 mL each of 0.824 M NH3 (aq) and 0.795 M acetic acid, the b) Write a thermochemical equation for the reaction of NHs and acetic acid using the calculated |neutralization from (a). Be sure to apply the appropriate phase labels and balancing. was found to be 4.52°C. The specific heat of the reaction mixture was 4.104Explanation / Answer
(1)
Assuming the density of water as 1 g/mL.
Heat lost by the hot water = m*C*dT = 100*4.187*(45-29.7) = 6406.11 J
Heat gained by cold water = m*C*dT = 100*4.184*(29.7-19.8) = 4142.16 J
So,
Heat gained by calorimeter = Heat difference in the transfer of energy from hot to cold water = 6406.11-4142.16 = 2263.95 J
Assuming Calorimeter constant as 'C', we have:
C*dT = 2263.95
Putting values:
C*(29.7-19.8) = 2263.95
Solving we get:
C = 228.68 (J/0C)
Hope this helps !