Question

A 91.70 g sample of metal was heated in a boiling water bath at 99.4 °C. The hot metal was then placed in a calorimeter, with heat capacity 39 J/K containing 45.0 g of water. Analysis of the thermogram showed the initial temperature to be 21.1°C, the final temperature to be 42.3°C.

4.91 g of LiCl was dissolved in 50.0 mL of water in the same calorimeter as in problem (above)

1. Analysis of the thermogram showed the initial temperature to be 22.9C, the final temperature to be 49.2C.

Was the dissolution of the ionic solid exothermic or endothermic?

Calculate the mass of the water

Calculate delta H solution

Calculate delta H ( - above H with degrees) solution (assume constant pressure of 1 bar)

Please post work because I want to learn this material.. not just get answers. Thanks

Answer 1

Dissolution is exothermic or endothermic?

The temperature rises from 22.9 C to 49.2 C, which clearly indicates that dissolution of the ionic solid is exothermic, since heat is released which increases the temperature.

Mass of water?

The normal density of water is 1 g/ ml. therefore the mass of water in which LiCl was dissolved = (1 g/ml)x (50 ml)

=50 g

Also the water present in the calorimeter is 45 g

Therefore total mass of the water =50 + 45

=95 g

The delta H_solution = (Heat required to raise the temperature of water) + (Heat required to raise the temperature of calorimeter)

Heat required to raise the temperature of water = m x c x (delta T); (c is specific heat of water =4.184 g/J-C)

=(95 g)x(4.184 g/J-C)x(49.2-22.9C)

=10453.7 J

Heat required to raise the temperature of calorimeter = (39 J/K)x(49.2-22.9C)

=1025.7 J

Therefore delta H_solution =10453.7+1025.7

=11479.4 J (liberated)   

(Kindly note that the heat is liberated and for exothermic reactions delta H is negative)

So, delta H_solution = -11479.4 J (-ve sign to indicate exothermic or liberation of heat)

Last Part bar above delta H is for standard molar enthalpy of the solution

Standard molar enthalpy of the solution = delta H_solution/ (moles of LiCl)

=11479.4 J/ (4.91g/42.5 g/mol) (Since Molar mass of LiCl is 42.5 g/mol)

=99363.4 J/mol