Additionally the masses and volumes of solution were

temperature rod had reached the same temperature as the solution before reaction. Additionally,
the masses and volumes of solution were duplicated for all reactions.

Data Tables:
Table 1:
Reaction 1 Reaction 2 Reaction 3
Mass of solid sodium
hydroxide (grams)
1.960 2.0225
Volume of deionized
water (mL)
100.00 50.00
Volume of 1.0 M HCl
(mL)
49.5 49.5
Volume of 1.0 M
NaOH (mL)
50.0
Initial temperature, Ti,
°C
25.6 24.8 23.7
Final temperature, Tf,
°C
30.1 36.0 30.5
Change in
temperature ∆T, °C
4.5 11.2 6.8
Table 2:
Reaction 4 Reaction 5 Reaction 6
Mass of solid sodium
hydroxide (grams)
1.960 2.0264
Volume of deionized
water (mL)
100.00 50.00
Volume of 1.0 M
CH3COOH (mL)
50.00 50.0
Volume of 1.0 M
NaOH (mL)
50.0
Initial temperature,
Ti, °C
25.6 24.3 23.5
Final temperature, Tf, 30.1 35.2 30.0

°C
Change in
temperature ∆T, °C
4.5 10.9 6.5
Calculations:
Reaction 1:
Mass of water =100.0g
ΔT = 4.5
o
C
Specific heat =
( ¿
o
Cg )
4.18 J / ¿
( ¿
o
Cg )( 100.00 g )∗ 4.5
o
C = 1880 J
q
water
( Joules )= 4.18 J / ¿
q
water
( kJ )= 1.88 kJ
Moles of NaOH =
0.049 mol
q
water
( kJ / mol )= 38.4 kJ / mol
q
reaction
( kJ / mol )=− 38.4 kJ / mol
Reaction 2:
Mass of water =99.5g
ΔT = 11.2 C
Specific heat =
( ¿
o
Cg )
4.18 J / ¿
( ¿
o
Cg )( 99.5 g )∗ 11.2 C = 4660 J
q
water
( Joules )= 4.18 J / ¿
q
water
( kJ )= 4.660 kJ
Moles of NaOH =
0.05056 mol
Moles HCl=
0.0500 mol
Limiting reactant: HCl
q
water
( kJ / mol )= 93.2 kJ / mol
q
reaction
( kJ / mol )=− 93.2 kJ / mol
Reaction 3:
Mass of water=99.5g
ΔT = 6.8 C
Specific heat=
( ¿
o
Cg )
4.18 J / ¿
( ¿
o
Cg )( 99.5 g )∗ 6.8
o
C = 2830 J
q
water
( Joules )= 4.18 J / ¿
q
water
( kJ )= 2.830 kJ

Moles NaOH=
0.0495 mol
Moles HCl=
0.0500 mol
Limiting reactant=NaOH
q
water
( kJ / mol )= 57.2 kJ / mol
q
reaction
( kJ / mol )=− 57.2 kJ / mol
Evaluating the Validity of Hess’s Law:
Measured
Δ H
3
=− 57.2 kJ
Value of ∆H for Reac tion 3 bas ed on Hess’s law = ∆H2 – ∆H1 =
− 93.2 kJ / mol −(− 38.4 kJ / mol )=− 54.8