CHEM1000 Lab 3 Report - Grade: 76 PDF

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CHEM 1000 Lab 3 Report (Enthalpies of Chemical Reactions), Fall 2020 Please write your name, student number and your signature on every page of work you submit. NOTE: This lab report must be an independently produced piece of work. Collaboration of any kind is not permitted and violates the Senate Policy on Academic Honesty.

PART 1 1. Complete the following table for the reaction of Mg with HCl (2 runs total). The ΔrH should be reported in kJ mol-1 of Mg. Show your work for the calculation of ΔrH in run 1.

RUN 1 →

Mass Mg (mg):

19.3 mg

RUN 2 →

Mass Mg (mg):

Mass: 19.3 mg / 1000 = 0.0193g

Moles Mg (mol):

Molar mass: 24.305g/mol

Mass: 19 mg / 1000 = 0.019g

Moles Mg (mol):

n = (0.0193 g) / (24.305g/mol)

Heat absorbed by water (kJ):

Reaction ΔrH (kJ mol-1):

ΔT=32.9°C – 20.4°C ΔT = 12.5°C Q = m*C* ΔT Q= 5g x 4.184j/g°C x 12.5°C Q = 261.5 J / 1000 Q = 0.262 kJ ΔrH = q/n ΔrH = 0.262 kJ / 0.000794 mols ΔrH = 329.97 ΔrH=-329.97kJ/mol

Molar mass: 24.305 g/mol n = 0.019g / 24.305 g/mol

n = 0.000794 mols ΔT (°C):

19mg

n = 0.000782 mols ΔT (°C):

Heat absorbed by water (kJ):

Reaction ΔrH (kJ mol-1):

ΔT=32.6°C – 20.6°C ΔT = 12°C Q = m*C* ΔT Q = 5g x 4.184j/g°C x 12.5°C Q = 251.04 J / 1000 Q = 0.251 kJ ΔrH = q/n ΔrH = 0.251 kJ / 0.000782 mols ΔrH = 320.97 ΔrH=-320.97kJ/mol

PART 2 2. Complete the following table for the reaction between HCl and NaOH (2 runs total). Run 1 ΔT (°C) ΔT=25.1°C – 20.4°C ΔT = 4.7°C

Run 2 ΔT (°C) ΔT=25.4°C – 20°C ΔT = 5.4°C

Average ΔT (°C) Average ΔT (°C) = (ΔT1 + ΔT2) / 2 Average ΔT (°C) = (4.7°C + 5.4°C) / 2 Average ΔT (°C) = 10.1°C / 2 Average ΔT (°C) = 5.1°C

Average Heat Absorbed by Water (kJ) Q = m*C* ΔT Q= 5g x 4.184j/g°C x 5.1°C Q = 106.7J / 1000

Number of moles of HCl (mol) # of moles of HCl = 0.0025 mols

Q = 0.107 kJ

Average ΔrH (kJ mol-1) ΔrH = q/n ΔrH = 0.107 kJ / 0.0025 mols ΔrH = 42.80kJ/mol ΔrH=-42.80kJ/mol

PART 3 3. Complete the following table for the reaction between NH 4Cl and NaOH (2 runs total). Run 1 ΔT (°C) ΔT=21.6°C – 21°C ΔT = 0.6°C

Run 2 ΔT (°C) ΔT=21.7°C – 19.5°C ΔT = 2.2°C

Average ΔT (°C) Average ΔT (°C) = (ΔT1 + ΔT2) / 2 Average ΔT (°C) = (0.6°C + 2.2°C) / 2 Average ΔT (°C) = 2.8°C / 2 Average ΔT (°C) = 1.4°C

Average Heat Absorbed by Water (kJ) Q = m*C* ΔT Q= 5g x 4.184j/g°C x 1.4°C Q = 29.29J / 1000 Q = 0.029 kJ

Number of moles of NH4Cl (mol) # of moles of NH4Cl = 0.0025 mols

Average ΔH (kJ mol-1) ΔrH = q/n ΔrH = 0.029 kJ / 0.0025 mols ΔrH = 11.60kJ/mol ΔrH=-11.60kJ/mol

PART 4 4. Complete the following table for the reaction between NH 3 and HCl (2 runs total). Run 1 ΔT (°C) ΔT=24.3°C – 19.7°C ΔT = 4.6°C

Run 2 ΔT (°C) ΔT=24.6°C – 21°C ΔT = 3.6°C

Average ΔT (°C) Average ΔT (°C) = (ΔT1 + ΔT2) / 2 Average ΔT (°C) = (4.6°C + 3.6°C) / 2 Average ΔT (°C) = 8.2°C / 2 Average ΔT (°C) = 4.1°C

Average Heat Absorbed by Water (kJ) Q = m*C* ΔT Q= 5g x 4.184j/g°C x 4.1°C Q = 85.77J / 1000

Number of moles of HCl (mol) # of moles of HCl = 0.0025 mols

Q = 0.086 kJ

Average ΔrH (kJ mol-1) ΔrH = q/n ΔrH = 0.086 kJ / 0.0025 mols ΔrH = 34.40kJ/mol ΔrH=-34.40kJ/mol

5. Based on your ΔrH values obtained from parts 2 and 3, predict the ΔrH for the reaction between NH 3 and HCl. HINT: You need to use Hess’s Law. Show your work. ΔrH (part 2) = -42.80 kJ/mol ΔrH (part 3) = -11.60 kJ/mol HCl + NH3 -> NH4Cl Δ rH =? HCl + NaOH -> NaCl + H2O ΔrH =-42.80 kJ/mol NH4Cl + NaOH -> NH3 +NaCl + H2O ΔrH = -11.60 kJ/mol H2O +NaCl + NH3 -> NH4Cl + NaOH ΔrH = 11.60 kJ/mol

= HCl + NH3 -> NH4Cl

ΔrH = -42.80 kJ/mol + 11.60 kJ/mol

HCl + NH3 -> NH4Cl ΔrH = -31.2 kJ/mol

6. For run 1 of the part 1 experiment (reaction of Mg with HCl), calculate the absolute error (ΔΔrH) of your calculated ΔrH value (in kJ mol-1). Use the lab manual to obtain individual errors from each experimental measurement. The lab manual appendix contains details of how error propagation values are calculated. Show your work. HINT: To start, calculate the absolute error in ΔT only (referred to as ΔΔT). Then, consider your calculation of ΔrH in question 1 (run 1) and write out a single algebraic expression for ΔrH in terms of: mMg, mH2O, CH2O, ΔT, MMg. Then you can consider the errors (if any) associated with each of those terms.

Errors Mmg = 0.00193g + 0.00005g MH2O = 5g + 0.05g ΔΔT = 12.5°C + 0.2°C

ΔMmg = (0.00005g / 0.0193g) = 0.0026 ΔMH2O = (0.05g / 5g) = 0.01 ΔΔT = (0.2 °C / 12.5°C) = 0.016 ΔMmg = 0 ΔCH2O = 0

ΔΔrH = ΔMmg + ΔMH2O + ΔΔT + ΔMmg + ΔCH2O *(-329.97 kJ/mol) = 0.0286 (-329.97kJ/mol) = 9.47

∴ The absolute error of ΔrH is -329.97kJ/mol + 9.47kJ/mol....