When 10.0 mL of a 2.0 M acetc acid solution at 20.6 degree C is reacted with 10.
ID: 1052812 • Letter: W
Question
When 10.0 mL of a 2.0 M acetc acid solution at 20.6 degree C is reacted with 10.0 mL of a 2.1 M NaOH solution at 21.0 degree C in a coffee cup calorimeter, the resulting temperature of the solution was determined to be 28.6 degree C. Calculate the heat of reactor for this acid-base neutralization. Assume the density and heat capacity of the solution to be the same as water. (d=1.00 g/mL and C_rho=4.l84J/g degree C) The calorimeter constant was 2.7 J/degree C. Select one: - 630 J - 650 J - 670 J - 2500 J -21 JExplanation / Answer
Initial solution temp = average of acid + NaOH temp since both presnet in same amounts
= (20.6+21) / 2 = 20.8
solution volume = 10+10 = 20 ml
solution mass = volume x density = 20 ml x 1g/ml = 20 g
Heat absorbed by solution = specific heat x mass x temperature change
= 4.184 J/gK x 20 g x ( 28.6-20.8)
= 652.7 J
Heat absorbed by calorimeter = specific heta of calorimeter x temp change
= 2.7 J/C x ( 28.6-20.8) C
= 21 J
Total heta absorbed by solution + calorimeter = 652+21 = 673 J = 670 J ( rounded value)
Thus Heta of reaction = -670 J ( -ve sign indicates energy released)
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