Question

When 10.0 mL of a 2.0 M acetic acid solution at 20.6ºC is reacted with 10.0 mL of a 2.1 M NaOH solution at 21.0ºC in a coffee cup calorimeter, the resulting temperature of the solution was determined to be 28.6ºC. Calculate the heat of reaction for this acid-base neutralization.   Assume the density and heat capacity of the solution to be the same as water. (d=1.00 g/mL and C_p=4.184J/gºC) The calorimeter constant was 2.7 J/ºC.

Answer 1

CH3COOH + NaOH = CH3COONa + H2O

HRxn = -Q/n

n = MV

Mole of acid = 2.0*10/1000 = 0.02 mole ACID

Moles of base= 2.1*10/1000 = 0.021 mole base

ratio is 1:1 acid:base so,acid is limiting agent

therefore

total volume of solution 20 ml which has mass = 20 g due to density of 1.00 g/ ml

heat liberted= heat given to acetic acid+ heat given to sodium hydroxide+ heat taken by calorimeter

= mass of acetic acid* specific heat* temperature difference + mass of sodium hydroxide* specific heat* temperature difference + calorimeter constant* temperature difference=

10*4.184*(28.6-20.6)+10*4.184*(28.6-21)+2.7*(28.6-21)

=334.72 J + 326.352 J + 21.06 J

=682.132 Joules

so..

HRxn = -682.132/(0.02) = 34106.6 J/ mole

= 34.11kJ/mol