The enthalpy of solution of a 50.0 g aqueous solution was measured in a calorime

Question

The enthalpy of solution of a 50.0 g aqueous solution was measured in a calorimeter. During the measurement, the temperature of the calorimeter contents (the solution) increased from 22.0 to 25.3 degree C. Assume the specific heat of the calorimeter contents is the same as the specific heat of water (4.184 J/g- degree C). Use Equation (11-5) to calculate Delta H_calorimeter during the dissolution. Use Equation (11-2) to calculate Delta H_solution from the value of Delta H_calorimeter obtained in the preceding step. From the sign of the value obtained in step b. determine whether the dissolution reaction is exothermic or endothermic. A 40.0 mL aliquot of 1.10 M HCl(aq) was mixed in a calorimeter with a 25.0 mL aliquot of 1.04 M NaOH(aq). The specific heat c for the resulting solution is 4.01 J/g degree C. The density of the solution is 1.02 g/mL. During the reaction the temperature of the solution increased from 23.2 degree C to 28.6 degree C. From the sum of the volumes of hydrochloric acid and aqueous sodium hydroxide, and the density of the solution, calculate the mass of the solution. Use Equation (11-5) to calculate Delta H_calorimeter. Use Equation (11-3) to calculate Delta H_neutralization from Delta H_calorimeter. Use the concentration and volume of the NaOH solution to calculate the number of moles of NaOH used in the reaction. Use the answers to parts c and d to calculate Delta H_neutralization in units of kJ/mol NaOH.

Explanation / Answer

1a) Heat absorbed bt the ccalorimeter =mass x specific heat x (T2 -T1)

= 50gx 4.184J/g x (3.3)

= 690.36 J

Thus delta H of calorimeter = 690.36J

1b) Delta H of the solution = - delta H of calorimeter

= -690.36 J

1c)Since the sign of delta H of solutioni is negative, the reaction is exothermic.

2a) mass of solution = volume x density

= (40 + 25)mL x 1.02g/mL = 66.35g

b) delta H of calorimeter = mass x specific heat x difference in temperature

= 66.35g x 4.01g/mL x (28.6 - 23.2)

= 1435.66J

c) delta H neutralization = -1435.66J

d) mmoles of acid = 40mLx1.1 M = 44

mmoles of base = 25mLx1.02 M = 26

Thus only 26 mmol of acid reacted with 26 mmoles of base.

For 26 mmoles reacted the heat liberated = 1435.66J

for 1 mole=1000mmoles heat liberated = 1000x 1435.66/26 = 55,217.7 j

= 55.217kJ/moles of NaOh

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