Molar Enthalpy of a Chemical Change

Purpose:

To use calorimetry to obtain an empirical value for the molar enthalpy of neutralization of sodium hydroxide by sulfuric acid.

Materials:

1.0mol/L sodium hydroxide solution1.0mol/L sulfuric acid solutionThermometerPolystyrene calorimeter2 100-mL graduated cylinders

Procedure:

Assumptions:

- No heat is lost to the the styrofoam cup.- No heat is lost to the air inside the styrofoam cup.

Observations:

Sulfuric Acid solution Sodium Hydroxide solution

Volume (ml) 30.0mL 49.5mL

Mass (g) 30.0g 49.5g

Init. Temperature (°C) 20.0°C 20.5°C

Final Temperature (°C) 28.5°C

∆ Temperature 8.5°C 8.0°C

Analysis:

30 mL H2SO4 50 mL NaOH

2NaOH + H2SO4 2H2O + Na2SO4

49.5mL NaOH 30.0mL H2SO4

c=n/v

n=cv

0.0495mol 0.0300mol

0.0495x1/2 0.03x 1/1

=0.02475 =0.03

Since 0.02475<0.03

NaOH is the Limiting Reagent

Sourabh Das 17/04/2023

b) Calculations:

(i) macid = 30.0g mbase = 49.5g

(ii) ∆Tacid = 8.5°C ∆Tbase = 8.0°C(iii) Q = |mc∆T|1 + |mc∆T|2

= (30.0 g×4.18 Jg ∙° C

×8.5 ° C)+(49.5 g×4.18 Jg∙ ° C

×8.0 °C ) = 1655.28 J + 1065.9 J = 2721.18 J = 2.72118 kJ

(iv) c = nv

n = c v

= (1.0 molL ∙0.0495 L) = 0.0495 mol

(v) ∆ H neu .= ∆ Hn

= (2.72118 kJ/ 0.0495 mol)

= 54.973 kJmol

≈ 55 kJmol

(2 s.f.)

Evaluation:

c) % difference:

¿−55−(−56 )∨ ¿¿−56∨¿×100%=1.8%¿

¿ error

d) Sources of error:

- Heat may be lost from the Styrofoam cup to the outside environment due to the hole on the lid for the thermometer therefore the cup is not a perfectly closed system (taking assumptions into account).

- During the transfer of the solutions from the graduated cylinder some droplets of solutions may remain in the original container making errors in measurements of volume.

- Solutions were exposed to matter external to the system while being transferred to the calorimeter which may lead to unaccounted change in temperature from our initial measured temperature.

e) Effects with different quantities of HsSO4 (aq):

(i) 100mL of HsSO4 (aq):Since the acid is in excess the total mass in the measurement would increase and the total measured change in temperature would decrease, such that they were inversely proportional leading to the same value for Q, and therefore the same calculated enthalpy.(ii) 20mL of HsSO4 (aq):The acid would be the limiting reagent as 20ML of HsSO4 (aq) would contain 0.02 mol which is less than 0.02475 mol of NaOH. Because of this, the measured temperature change would be lower, and the total mass would be about 70g which is also lower. This would

Sourabh Das 17/04/2023

lead to a lower value of Q. However since the total number of moles reacted is also less, the calculated enthalpy would still arrive at the same constant of -56 kJ/mol.