Enthalpy Change of a Displacement Reaction

Alan Williams
Enthalpy Change of a Displacement Reaction

Data Collection and Processing

Quantitative Data:

Table 1: Raw Data showing Temperature during 9 minutes (Trail1,2)
Time (sec) (±1.0) Temperature (°C)(±0.5) Trial 1 Temperature (°C)(±0.5) Trial 2
0 19.0 20.0
30 19.0 20.0
60 19.0 20.0
90 19.0 20.0
120 19.0 20.0
150 19.0 20.0
180 40.0 30.0
210 45.0 52.0
240 46.0 53.5
270 47.0 54.0
300 48.0 53.5
330 48.0 53.0
360 48.0 52.5
390 47.5 52.0
420 47.0 52.0
450 46.5 51.5
480 46.0 51.0
510 45.5 50.0
540 45.0 49.5

Qualitative Observation:

CuSO4 solution Blue aqueous
Zn powder Black powder
The copper sulfate solution had a blue colour at room temperature, but as the reaction progressed and the zinc powder was added, the warmed up solution became a dark gray colour with red-brown deposits in it.

Calculations
CuSO4 solution specific heat capacity x Temperature rise x mass of copper sulfate solution = Heat Evolved (kJ)

Trial 1
Temperature rise = max temperature rise- initial temperature
= 4.2 x (56- 19) x 25.0g   = 3.89 kJ

Moles of CuSO4 solution
Molarity= moles/ volume 1 M= n/0.025 dm3 Moles= 0.025 of CuSO4 solution

Enthalpy of this reaction
-3.89/0.025 = -155.6 kJ/mol

Therefore:
Zn(s) + Cu 2+ (aq)Cu (s) + Zn 2+ (aq)   ΔH = -155.6 kJ/mol

Trial 2
Temperature rise = max temperature rise- initial temperature
= 4.2 x (62- 20) x 25.0g   = 4.41kJ

Moles of CuSO4 solution
Moles= 0.025 of CuSO4solution

Enthalpy of this reaction
-4.41/0.025 = -176.4 kJ/mol

Therefore:
Zn(s) + Cu 2+ (aq)Cu (s) + Zn 2+ (aq)
ΔH=-155.6 kJ/mol
ΔH=-176.4 kJ/mol

Average enthalpy change
(-155.6 kJ/mol+ -176.4 kJ/mol)/2 =-166.0kJ

The accepted value in the data book: -218kJ/mol

Error Calculation

Variables Margin of Error Error Percentage (%)
Temperature (°C) ±1.0 Trial 1
(1.0/(56-19)) x 100
= 2.70%

Trial 2
(1.0/(62-20)) x 100
= 2.38%
Average: 2.54%
Time (secs) ±1.0 Min: (1.0/30) x 100
= 3.33%

Max: (1.0/540)...