When a neutralization reaction was carried out using 100.0 mL of 0.7550M NH3 wat

Question

When a neutralization reaction was carried out using 100.0 mL of 0.7550M NH3 water and 100.0 mL of 0.7645M acetic acid, AT was found to be 4.66 degree C. 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 3.60 JK-1. Calculate the AHneulzn for the reaction of NH3 and acetic acid. In the experiment in the previous question, what if it was found that the thermometer read 0.5 degree C too high? What effect would this have on the Delta Hneutzn calculated in the previous question? Raise the Delta Hneutzn Lower the Delta Hneutzn Have no effect on Delta Hneutzn In the experiment in the previous question, what if it was found that there was an error in the change in temperature and the AT should have been lower? What effect would this have on the Delta Hneutzn calculated in the previous question? Raise the Delta Hneutzn Lower the Delta Hneutzn Have no effect on Delta Hneutzn 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? Larger Smaller Suppose you did your experiment in a test tube without insulation. Would you expect the Delta Hneutzn for the test tube alone to be larger, smaller or the same as what you found with the insulated test tube? Larger Smaller Same

Explanation / Answer

6 )

q (heat changed) = m*s*DT + DT*Ccal

mass of solution = 100+100 = 200*1.03 = 206 grams

S = 4.104 j/g.k

DT = 4.66 c

Ccal = 3.6 j/k

q = 206*4.104*4.66 + 3.6*4.66 = 3.956 kj

No of moles of NH3 = 0.1*0.755 = 0.0755 Mol

DHneu = -3.956/0.0755 = -52.4 kj/mol

7) as the thermometer reading is increased ,than actual value,

is increased

answer : A) raise the DHneu

8) B) lower the DHneu

9 ) same

10 ) same

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