kJmol 42302 kJ mol \u0394H1 REACTION 2 Mass of water 10 g mL 5000 mL 5000 mL 20040 g

# Kjmol 42302 kj mol δh1 reaction 2 mass of water 10 g

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(kJ/mol) = - 42.302 kJ mol = ΔH1 REACTION 2 Mass of water = 1.0 g mL × ( 50.00 mL + 50.00 mL )+ 2.0040 g = 102.0040 g ΔT = T f – T i = 35.4 - 24.4 = 11.0 °C Specific heat = 4.18 Joules gram q water (Joules) = 102.0040 4.18 Joules gram × 11.0 = 4690.1439 Joules q water (kJ) = 4690.1439 Joules× 1 kJ 1000 joules = 4.690 kJ Moles of NaOH = 2.0040 1 mol 39.99 g = 0.0501 moles Moles of HCl = 1 M HCl× ( 50.0000 mL× 1 L 1000 mL )= 0.0500 moles q water (kJ/mol) = 4.690 kJ 0.0500 moles = 93.800 kJ mol q reaction (kJ/mol) = - 93.8 kJ mol = ΔH2 REACTION 3 Mass of water = 1.0 g mL × ( 50.00 mL + 50.00 mL ) = 100.00 g ΔT = T f – T i = 29.9 - 23.1 = 6.8 °C Specific heat = 4.18 Joules gram q water (Joules) = 100.00 4.18 Joules gram × 6.8 = 2842.4 Joules
q water (kJ) = 2842.4 Joules× 1 kJ 1000 joules = 2.842 kJ Moles of NaOH = 1 M NaOH × ( 50.0000 mL× 1 L 1000 mL )= 0.0500 moles Moles of HCl = 1 M HCl× ( 50.0000 mL× 1 L 1000 mL )= 0.0500 moles q water (kJ/mol) = 2.842 kJ 0.0500 moles = 56.840 kJ mol q reaction (kJ/mol) = - 56.840 kJ mol = ΔH3 % error = ( 42.302 kJ mol +− 56.840 kJ mol ) −− 93.800 kJ mol 93.800 kJ mol × 100 = 5.69 REACTION 4 Mass of water = 1.0 g mL × 100.00 mL + 2.0151 g = 102.0151 g ΔT = T f – T i = 29.2 - 24.2 = 5.0 °C Specific heat = 4.18 Joules gram q water (Joules) = 102.0151 4.18 Joules gram × 5.0 = 2132.1156 Joules q water (kJ) = 2132.1156 Joules× 1 kJ 1000 joules = 2.132 kJ Moles of NaOH = 2.0151 1 mol 39.99 g = 0.0504 moles q water (kJ/mol) = 2.132 kJ 0.0504 moles = 42.302 kJ mol q reaction (kJ/mol) = - 42.302 kJ mol = ΔH4 REACTION 5 Mass of water = 1.0 g mL × ( 50.00 mL + 50.00 mL )+ 2.0336 g = 102.0336 g ΔT = T f – T i = 35.9 - 24.7 = 11.2 °C Specific heat = 4.18 Joules gram q water (Joules) = 102.0336

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