eaction kJmol 4348 kJmoles H 2 Reaction 3 NaOHaq HClaq NaClaq H2Ol T Tf Ti 625C

# Eaction kjmol 4348 kjmoles h 2 reaction 3 naohaq

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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

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• Summer '19
• Sodium, Sodium hydroxide