A 48.53 mL volume of 1.00 M HCl was mixed with 47.70 mL of 2.00 M...
A 48.53 mL volume of 1.00 M HCl was mixed with 47.70 mL of 2.00 M NaOH in a coffee cup calorimeter (with calorimeter constant = 26.0 J/°C) at 21.43 °C. The final temperature of the aqueous solution after the reaction was 29.71 °C. Assuming that them heat capacity of the solution is 4.18 J/g/°C, calculate the following: a. The total mass of aqueous solution inside the calorimeter (dsoln = 1.00 g/mL) g b. The change in temperature (∆T) of the aqueous solution °C c. The heat released by the neutralization reaction (this should be a positive number) J d. The moles of HCl reacted mol e. The enthalpy change (∆H) for the neutralization in kJ/mol HCl ( this should be a negative number) kJ/mol HCl
Homework Answers
a) total volume of solution = 48.53 + 47.70 mL =96.23 ml
sinc edensity = 1g/mL,
mass of solution in calorimeter = 96.23 g
b)
change in temperature DT = Tfinal- Tinitial = 29.71-21.43 =8.28C
c) -Heat released by the neutralization = heat gained by solution + heat gained by calorimeter
= mass x sp. heat of water x DT + calorimeter constant x DT
= [96.23g x4.184J/g.C x8.28 ] + 26J/C x 8.28C
=3549.03 J
d)moles of HCl reacted
HCl + NaOH -------------------------> NaCl + H2O
48.53x1 47.7x2=95.4 0 0 intital mmoles
0 46.87 48.53 - after reaction
thus all hCl is reacted
So moles of HCl reacted = 48.53/1000 = 0.04853 mol
e) when 0.04853 moll of Hcl reacted heat liberated = -3549.03J
If 1 mol reacted = -3549.03 /0.04853 =-73130J
= -73.13 kJ/mol
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