4 Closed the lid of the calorimeter and stirred the solution until the

# 4 closed the lid of the calorimeter and stirred the

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4. Closed the lid of the calorimeter and stirred the solution until the temperature remained constant for at least 30 seconds. Recorded the temperature change. 5. Calculated the enthalpy change. Data: Table A (Unknown Metal) Mass of Metal Mass of Water Initial Temp. of Initial Temp. of Final Temp. of Water and Specific Heat of Water Metal Metal Metal Trial 1 6.205 g 99.002 g 21.5 99.7 22.2 0.603 J Trial 2 6.330 g 98.944 g 21.7 99.7 22.2 0.422 J Trial 3 2.186g 99.278 g 22.1 99.8 22.4 0.737 J Table B (Sodium Hydroxide) Mass of Sodium Hydroxide Mass of Water (Initial) Initial Temperature of Water Final Temperature of Solution Change in Temperature Change in Enthalpy H) (? Trial 1 4.003 g 99.748 g 21.9 31.5 9.6 -42 kJ/mol Trial 2 4.027 g 99.564 g 21.8 31.4 9.6 -41 kJ/mol Trial 3 4.019 g 99.002 g 21.8 31.6 9.8 -42 kJ/mol Table C (Magnesium Metal) Solid Reactants Mass Of Solid Reactant Mass of HCl Initial Temp. of HCl Final Temp. of Reaction Temperature Change Change in Enthalpy H) (? Magnesium 1.214g 99.883 g 23.6 60.8 37.2 -311.2 kJ/mol Magnesium Oxide 2.017g 100.520g 21.4 32.8 11.4 -95.80 kJ/mol Calculations 1) Determining the Specific Heat of the Unknown Metal : Trial 1: Q = mC Δ T , Q metal = -Q water (6.205g unknown )(C)(-77.5C°) = - (99.002g H O )(4.184 J )(0.7C°) -480.8875C = -289.957058 C = -239.957058 / -480.8875 C = 0.603 J Trial 2 : Q = mC Δ T , Q metal = -Q water (6.330g unknown )(C)(-77.5C°) = - (98.944g H O )(4.184 J )(0.5C°) -490.575C = -206.990848 C = -206.990848/-490.575 C = 0.422 J Trial 3 : Q = mC Δ T , Q metal = -Q water (2.186g unknown )(C)(-77.4C°) = - (99.278g H O )(4.184 J )(0.3C°) -169.1964C = -124.613746 C = -124.613746 / -169.1964 C = 0.737 J Average Specific Heat of Unknown : ( 0.603 J + 0.422 J + 0.737 J ) / 3 = 0.587 J 2) Determining the Enthalpy of the Dissolution of Sodium Hydroxide : Trial 1 : Q = mC Δ T , Q metal = -Q water -(4.003g NaOH + 99.748 gH2O )(4.184 J g·C )(9.6) = -4167.304166 J -4167.304166 J / 1000 = -4.167304166 kJ ΔH = 4.167304166 kJ 4.003 g NaOH x 39.998 gNaOH 1 molNaOH = -41.63972821 ≈ -42 kJ / mol l Trial 2 : Q = mC Δ T , Q metal = -Q water -(4.027g NaOH + 99.564 gH2O )(4.184 J g·C )(9.6) = -4160.877542J -4160.877542J/1000 = -4.160877542 kJ ΔH = 4.160877542 kJ 4.027 g NaOH x 39.998 gNaOH 1 molNaOH = -41.32773278 kJ/mol ≈ -41 kJ / mol Trial 3 : Q = mC Δ T , Q metal = -Q water -(4.019g NaOH + 99.002 gH2O )(4.184 J g·C )(9.8) = -4224.190667J -4224.190667J / 1000 = -4.224190667 kJ ΔH = 4.224190667 kJ 4.019 g NaOH x 39.998 gNaOH 1 molNaOH = -42.04010408kJ/mol ≈ -42 kJ / mol Average Calculation of the dissolution of the NaOH: (-42 kJ/mol + -41 kJ/mol + -42 kJ/mol) / 3 = -41.6333333 kJ/mol ≈ -42 kJ / mol 3) Determine the enthalpy change of the reaction of magnesium metal and oxygen gas using Hess’s law : For Reaction 1: Mg(s) + 2HCl(aq) → MgCl 2 (aq) +H 2 (g) Since 1.0 M HCl is mostly water we can assume the HCl solution to have the properties of water: -Q water = −( 99.883 g )( 4.184 J )( 37.2 )=− 15546.3 J ≈ 15.546 kJ ΔH = 15.546 kJ 1.214 gMg x 24.30 gMg 1 mol = -311.2 kJ For Reaction 2: MgO(s) +2HCl (aq) → MgCl 2 (aq) + H 2 O (l) Since 1.0 M HCl is mostly water we can assume the HCl solution to have the properties of water: - Q water = -(100.520g)( 4.184 J )( 11.4 C ° ) = -4794.562752 J 4.79456 kJ ΔH = 4.79456 kJ 2.017 gMgO x 40.3 gMgO 1 mol = -95.79611701 kJ / mol -95.80 kJ For Reaction 3: Using Standard Enthalpy of Formation Given Equation : H 2 (g ) + O 2 (g) → H 2 O ( l) Balanced Equation : 2H 2 (g) + O 2 (g) → 2H 2 O ( l) ΔH of H 2 = 0 kJ / mol ΔH of O 2 = 0 kJ / mol ΔH = -285.8 kJ ΔH reaction = 2(-285.8) - 0 = -571.6 kJ / mol Hess’s Law Calculations for reaction of Magnesium and Oxygen Eq 1 : Mg(s) + 2HCl(aq) → MgCl 2 (aq) +H 2 (g) ΔH = -311.2 kJ Eq 2 : MgO(s) +2HCl (aq) → MgCl  #### You've reached the end of your free preview.

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