DATA TABLE
| Sample | Mass of water | Initial tempOf water | Final TempOf water | Change In tempOf water | Energy change use mc△t | Mass of salt used (g) | Enthalpy of solution kJ/g | Enthalpy of solutionkJ/mol |
| NaOH | 75g | 21.4 | 40.4 | 190C | – 5.9565kJ | 6.28g | -0.93395 | -0.147KJ/mol |
| KCl | 75g | 21.3 | 16.3 | 50C | +1.5675kJ | 5.98g | + 0.2676KJ | +0.214KJ/MOL |
Mass of the solution = 1g/cm3 x75cm3 = 75g
Change in temp of solution NaOH = 40.4 -21.4 = 190C
Change in temp of solution of KCl = 21.3 – 16.3 = 5 0C
Enthalpy of solution = mass of solid x specific heat capacity x change in temperature
△H= M x Q x △T
Enthalpy of sodium hydroxide = – (75g x 4.18J/g/K x 19K)
= – 5956.5J
Using the same formula for calculating enthalpy we can calculate the
molar enthalpy of potassium chloride = 75g x 4.18J/g/ K x 5k
= +1567.5J
The enthalpy in kJ=
1KJ = 1000J
For sodium hydroxide = -5956.5J/1000
= -5.9565KJ
For potassium chloride = +1567.5J/1000
= + 1.5675KJ
Moles = mass / molar mass
Mole of sodium hydroxide = 6.28/39.997
= 0.1570 moles
Moles of potassium chloride = 5.98/ 74.5513
=0.00802 moles.
Molar enthalpy of the solutions will be:
Molar enthalpy = enthalpy / mole
For sodium hydroxide = -0.147KJ/mol
For potassium chloride = +0.214KJ/MOL
1. Calculate the percent error for both salt enthalpy of solution.
Formula and accepted values are provided.
% error = ( accepted value – experimental value )/ accepted value
NaOH= ( -44.5- (-0.147) -44.5 x 100= 99.6%
The error for sodium hydroxide is 99.6%
KCl = (17.55- (0.214)/17.55x 100= 98.78%
2. Suggest three sources of error in this experiment, not including human errors.
The heat produce may be gained or lost by the apparatus during the experiment, also the calorimetry insulation may be inefficient.
The third source of error may be because of incomplete reaction where the results are taken before reaction is complete.
3. If some heats were transferred to the air or Styrofoam cup, would your
calculated enthalpy of solution of the salt be too high or too low? Explain
your answer
it would be too low because the projected temperature change would be too small due to heat loss.
4. If some salts were accidentally spilled as it was transferred from the balance
to the cup, would your calculated enthalpy of solution of the salt be too higher too low? Explain your answer.
the mass of salt added to the solution would decrease and lower than the intended amount. If there was less solute than the intended amount, then the is released or absorbed would be lower than the calculated amount. The accurate mass of solute is important as it give accurate results.
