Question

1. When 4.203g of NaOH is added to 100.0mL of water, the temperature of the solution increases by 3.0C. Assume the density of the solution to be 1g/mL and the specific heat of capacity of the solution to be 4.184 J/gK.

a) Calculate the amount of heat absorbed by the solution.

b) Find the enthalpy change for the dissolution of NaOH.

c) Find the enthalpy change per mole of NaOH.

Answer 1

When 4.203g of NaOH is added to 100.0mL of water, the temperature of the solution increases by 3.0C. Assume the density of the solution to be 1g/mL and the specific heat of capacity of the solution to be 4.184 J/gK.

1. Calculate the amount of heat absorbed by the solution.

Heat absorbed = q = cmT

Q= Heat transferred

C = specific heat of solution

M= mass of solution

T= temperature change

4.184 (100) (3.0)

=1,255.2 J

b) Find the enthalpy change for the dissolution of NaOH.

We know the heat capacity of water is 1 calorie per gram per degree C, or
4.184 joules per calorie per degree C (it is the same thing; 1 cal. = 4.184 J.)

If the calorimeter absorbs negligible heat, we know if we have 100 ml. water
its mass is 1 gram per ml. so we have 100 grams of water.

The temperature rose to 3°C so the sodium hydroxide released

Now we need the number of moles of NaOH we have. The molecular weight
is 40 and we have 4.203 grams, so that is 4.203/40 or 0.1050 mole which gave
off 300.00 calories of heat.

So the molar enthalpy change is 300/0.1050 or 2857 calories per mole (28.57 kcal. per mole)

If you want it in joules, multiply by 4.184 and obtain 119.541 KJ per mole.

C)Find the enthalpy change per mole of NaOH

ΔH = -q/1000 ÷ n(solute) = -1255.2/1000 ÷ 0.1050(moles of NaOH) = -12 kJ mol^-1