Eaction kjmol 4348 kjmoles h 2 reaction 3 naohaq

eaction
(kJ/mol) = -43.48 kJ/moles = ∆H
2
Reaction 3:
NaOH(aq) + HCl(aq) –––> NaCl(aq) + H2O(l)
∆T = Tf – Ti = 6.25°C
Specific heat = 4.18 J/g*C
qwater (Joules) = (4.18 J/g*C)(99.1 grams)(6.25°C)= 2589 Joules
qwater (kJ) = 2.589 Joules
Moles of NaOH = (0.0495 L)(1.00 mol NaOH/1 L)(1 mol HO/1 mol NaOH)= 0.0495 moles
Moles of HCl = (0.0496 L)(1.00 mol NaOH/1 L)(1 mol H
2
O/1 mol NaOH)=0.0496 moles
Limiting Reactant = neither; both were present in equal amount
q
water
(kJ/mol) = 2.589 Joules/0.0495mol = 52.30 kJ/moles
q
reaction
(kJ/mol) = -52.30 kJ/moles = ∆H3
Value of ∆H for Reaction 3 based on Hess’s law = ∆H2 – ∆H1 = -6.01 kJ/moles
Percent error in the calculated value compared to the measured value:
(∆H2 – ∆H1)/∆H3 = (-6.01/52.30)*100 = 11.5%
RXN 2: NaOH(s) + HCl(aq) –––> NaCl(aq) + H
2
O(l) ∆H= -37.47 kJ/mol
RXN 1: + NaOH(s) –––> NaOH(aq) ∆H= -43.48 kJ/mol

____________________________________________________
NaOH(aq) + HCl(aq) –––> NaCl(aq) + H2O(l) ∆H= -80.95 kJ/mol
Na + + OH - + H + + Cl - ––> Na + + Cl - + H
2
O
OH - + H + ––>H
2
O
NaOH(s) + OH- + H + ––> Na + + OH - + H
2
O
Final: NaOH(s) + H + (aq) ––> Na + (aq) + H
2
O ∆H = -80.95 kJ/mol
Reaction 4:
(same as reaction one)
Mass of water = 99.4 grams
∆T = Tf – Ti = 4.6°C
Specific heat = 4.18 water
q
water
(Joules) = (4.18 J/g*C)(99.4 mL)(4.6 ° C) = 1911 Joules
q
water
(kJ) = 1.911 kJ
Moles of NaOH = (2.03 grams)(1 mole/39.99 grams) = 0.051 moles
q
water
(kJ/mol) = 1.911/0.051 = kJ/mol
q
reaction
(kJ/mol) = -37.47 kJ/mol = ∆H
1
Reaction 5:
NaOH(s) + CH
3
COOH(aq) –––> NaCH
3
COO(aq) + H
2
O(l)
Mass of water: 49.60 grams
∆T = Tf – Ti = 10.6°C
qwater (Joules) = (4.18 J/g*C)(49.60 mL)(10.6°C) = 2211 Joules
qwater (kJ) = 2.211 kJ
Moles of HCl = (49.00 mL) (1.18 grams/1mL)(1 mole/36.46 grams)= 1.59 moles
Moles of CH
3
COOH = (49.00 mL) (1.05 grams/mL)(1 mole/60.05 grams)= 0.857 moles
Limiting Reactant: CH
3
COOH
q
water
(kJ/mol) = 2.211 kJ/0.857 moles =2.58 kJ/moles
qr
eaction
(kJ/mol) = -2.58 kJ/moles = ∆H
5
Reaction 6:
NaOH(aq) + CH
3
COOH(aq) –––> NaCH
3
COO(aq) + H
2
O(l)
Mass of water = (49.2 mL+48.9 mL)(1.00 g/mL)= 98.1 grams