1 -When a neutralization reaction was carried out using 100.0 mL of 0.7715M NH3

Question

1 -When a neutralization reaction was carried out using 100.0 mL of 0.7715M NH3 water and 100.0 mL of 0.7420M acetic acid, T was found to be 4.76ºC. The specific heat of the reaction mixture was 4.104 J g-1 K-1 and its density was 1.03 g mL-1. The calorimeter constant was 3.24 J K-1. Calculate the Hneutzn for the reaction of NH3 and acetic acid.

2- In the experiment in the previous question, what if it was found that there was an error in the change in temperature and the T should have been lower? What effect would this have on the Hneutzn calculated in the previous question?

A) Raise the Hneutzn

B) Lower the Hneutzn

C) Have no effect on Hneutzn

Explanation / Answer

we know that

moles = molarity x volume (L)

so

moles of acetic acid = 0.742 x 0.1 = 0.0742

moles of NH3 = 0.7715 x 0.1 = 0.07715

we can see that

acetic acid is the limiting reagent

now

total volume = volume of NH3 + volume of acetic acid

total volume = 100 + 100

total volume = 200 ml

now

we know that

mass = density x volume

given

density of the solution = 1.03

so

mass of solution = 1.03 x 200

mass of solution = 206 g

now

Heat = heat of the mixture + heat of the calorimeter

so

Heat = ( m x s x dT ) of mixture + ( C x dT ) of calorimeter

Heat = (206 x 4.104 x 4.76 ) + ( 3.24 x 4.76)

Heat = 4039.64 J

this is for 0.0742 moles of acetic acid

heat for 1 mole = 4039.64 / 0.0742

heat for 1 mole = 54442.59 J

heat for 1 mole = 54.44 kJ

so

the heat of neutralization for the reaction is 54.44 kJ

2)

if dT should have been lower , then lower the dH neutzn

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.