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

(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 g× 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 g× 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 g× 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 g× 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 g× 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