1 -When a neutralization reaction was carried out using 100.0 mL of 0.7715M NH3
ID: 918051 • Letter: 1
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 previous experiment, Suppose you did your experiment in a test tube without insulation. Would you expect the calorimeter constant for the test tube alone to be larger, smaller or the same as what you found with the insulated test tube?
A) Larger
B) Smaller
C) Same
Explanation / Answer
mass of solution = 200*1.03 = 206 grams.
specific heat of solution = 4.104 j/g.k
DT = 4.76 C
q absorbed = 206*4.104*4.76+ 3.24*4.76
= 4.04 Kj.
no of moles of NH3 = 0.1*0.7420 = 0.0742 mol
DHrxn = -q/n = -4.04/0.0742 = -54.45 kj/mol
2. c.same
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