calorimetry lab Trial 1 Trial 2 Trial 3 Trial 4 initial temp NaOH 22.3 C 23.1 22
ID: 881818 • Letter: C
Question
calorimetry lab
Trial 1 Trial 2 Trial 3 Trial 4
initial temp NaOH 22.3 C 23.1 22.0 22.1
initial temp acid 22.0 c 23.0 22.5 22.7
final temp 38.2 C 36.0 37.7 39.0
delta T 15.09 12.9 15.7 16.9
mass of solution 76.021g 76.021 75 78
moles of limiting reagent 0.098 0.09 0.09 0.096
Find
1, Delta H
2, Average delta H
3, Standard divation delta H
4 % erroe delta H
Explanation / Answer
Trial-1
First let us calculate the heat (Q) for the reaction
The amount of heat released corresponds to:
q = (76.021 g)(4.184 J/g*C)(38.2 - 22.0 C) = 5152.76 J
Since the reaction takes place in a calorimeter, one can assume pressure is constant. Under these conditions one can also assume the Q of the reaction = the change in enthalpy (Delta H)
Hence for 0.098 moles Delta H = 5152.76 x 0.098 = 504.97 J
Delta H1 = = 504.97 J
Trial-2
Similarly, if calculated
q = (76.021 g)(4.184 J/g*C)(36 - 23 C) = 4134.93 J
For 0.09 moles Delta H = 4134.93 x 0.09 = 372.14 J
Delta H2 = = 372.14 J
Trial-3
q = (75 g)(4.184 J/g*C)(37.7 - 22.5 C) = 4769.76 J
For 0.09 moles Delta H = 4769.76 x 0.09 = 429.27 J
Delta H3 = = 429.27 J
Trial-4
q = (78 g)(4.184 J/g*C)(39 - 22.7 C) = 5319.53 J
For 0.096 moles Delta H = 5319.53 x 0.096 = 510.67 J
Delta H4 = = 510.67 J
2) Average delta H : (504.97 J + 372.14 J+ 429.27 J + 510.67 J)/4
= 454.26 J
3, Standard divation delta H : 66.13
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