When 1.095 grams of NaOH was dissolved in a coffee cup calorimeter containing 150.00 grams of water, the temperature of the solution changed from 23.50 degree C to 25.32 degree C. a) Calculate the heat of solution, Delta H_soln.
b) Calculate the Delta H_soln/mole of NaOH. The molar mass of NaOH is 40.00 grams/mole.
c) Is the heat of solution exothermic or endothermic?

1. HEAT OF SOLUTION IS 1149.47 J

2. HEAT OF SOLUTION PER MOLE OF NaOH IS 41989.77 J/MOL OR 41.989 KJ/MOL

3. THE REACTION IS EXOTHERMIC

Explanation:

Mass of water = 150 g

Mass of NaOH = 1.095 g

Change in temperature = 25.32 - 23.50 = 1.82 C

Total mass of the mixture = 150 + 1.095 = 151.095 g

Specific heat capacity of water = 4.18 J/ g C

So to calculate the heat of solution, we have:

1. Heat of solution = mass * specific heat * change in temperature

Heat of solution = 151.095 * 4.18 * 1.82

Heat of solution = 1149.47 J

The heat of solution is therefore 1149.47 J or 1.15 kJ

2. Heat of solution per mole

1.095 g of NaOH produces 1149.47 J of heat

1 mole of NaOH contains 40 g

In other words:

1.095 g = 1149.47 J

40 g = x J

xJ = 40 * 1149.47 / 1.095

xJ = -41989.77 J

The heat of solution per mole of NaOH is -41.989 kJ / mol

3. The reaction is exothermic that is heat is evolved from the reaction. This is indicated by the negative sign.