Question

When a neutralization reaction was carried out using 100.0 mL of 0.7890 M NH₃ and 100.0 mL of 0.7940 M acetic acid, ∆T was found to be 4.76^oC. The specific heat of the reaction mixture was 4.104 J g^-1 K^-1 and its density was 1.03 g/mL. The calorimeter constant was 3.36 J K^-1. Show all work below for the following calculations:

a.) heat absorbed by the reaction mixture

b.) heat absorbed by the calorimeter

c.) total heat absorbed

d.) number of moles of acetic acid

e.) ∆H_{neutralization}

Answer 1

HC2H3O2 (aq) + NH3 (aq)   NH4+ (aq) + C2H3O2-(aq)

total volume of reaction mixture = 200 mL

density of mixture = 1.03 g/mL

mass of reaction mixture = 1.03*200

= 206 g

T = 4.76 oC

(a). heat absorbed by the reaction -

q1 = mcT

= 206*4.104*4.76

= 4.024 kJ

(b). heat absorbed by the calorimeter

q_cal = cT

= 3.36*4.76

= 15.99 J

c.) total heat absorbed

q = q1 + q_cal

= 4024.22 + 15.99

= 4040.21 J

= 4.04 kJ

d.) number of moles of acetic acid

moles of acetic acid = 0.100*0.7940

= 0.0794 moles