Question

1. In one of the General Chemistry experiments, you used a coffee-cup calorimeter to measure the heat of neutralization of selected acid-base reactions.
   The calorimeter consisted of two nested Styrofoam cups with a cardboard lid. A temperature probe was lowered in the solution through a hole in the lid.

   1. Is a coffee-cup calorimeter a constant-volume or a constant-pressure device?

   2. Is a coffee-cup calorimeter adiabatic?

   The coffee-cup calorimeter was calibrated by using the neutralization reaction between sodium hydroxide and hydrochloric acid. The heat output for this reaction is known to be 58 kJ/mol.

c. When 45.0 mL of 3.0 M NaOH are mixed with 60.0 mL of 3.0 M HCl, the temperature of the solution rises by 16.5°C. Calculate the calorimeter constant.

The specific heat capacity of the solution is 4.184 J/°C.g and the density of the solution is 1.00 g/mL.
The same coffee-cup calorimeter was then used to determine the heat of neutralization for an unknown acid-base pair. When 30.0 mL of a 2.78M solution of the unknown base were mixed with 50.0 mL of a 5.00 M solution of the unknown acid, the temperature of the solution rose from 23.5°C to 35.1°C.

d. Calculate the molar heat of neutralization for this acid-base reaction. Assume the same specific heat capacity and density of solution and calorimeter constant as in c).

Just need the solution for part d. Please show units!

Answer 1

c)

no of mol of NaOH taken = 45*3/1000 = 0.135 mole

no of mol of HCl taken = 60*3/1000 = 0.18 mole

limiting reactant = NaOH

amount of heat released in the reaction = 58*0.135 = 7.83 kj

heat released = heat gained by cold water + calorimeter

   7.83*10^3 = m*S*DT + C*DT

m = mass of reaction solution = 105 g   ( density of solution = 1 g/ml)

s = specific heat of solution = 4.184 j/g.c

DT = 16.5 c

C = heat capacity of calorimeter = ?

DT = 16.5 c

7.83*10^3 = 105*4.184*16.5 + c*16.5

C = heat capacity of calorimeter = 35.22 j/c

d)

no of mol of base taken = 30*2.78/1000 = 0.0834 mole

no of mol of acid taken = 50*5/1000 = 0.25 mole

limiting reactant = base

heat released(q) = heat gained by cold water + calorimeter

   q = m*S*DT + C*DT

m = mass of reaction solution = 80 g   ( density of solution = 1 g/ml)

s = specific heat of solution = 4.184 j/g.c

DT = 35.1-23.5 = 11.6 c

C = heat capacity of calorimeter = 35.22 j/c

DT = 35.1-23.5 = 11.6 c

q = 80*4.184*11.6 + 35.22*11.6

q = 4.29 kj

molar heat of neutralization (DHneu) = -q/n

                                     = -4.29/0.0834

                                     = -51.43 kj/mol