Find the difference between the value of x2 and the sum of x1 plus x3.; let x4 be equivalent to the sum of x1 and x3 and let x5 be the difference between x2 and x5. This calculation scheme is shown below. Make notes about any similarities or differences between the values in your notes. x4 = x1 + x3 x5 = x2 – x4
Find the difference between the value of x2 and the sum of x1 plus x3.; let x4 be equivalent to the sum of x1 and x3 and let x5 be the difference between x2 and x5. This calculation scheme is shown below. Make notes about any similarities or differences between the values in your notes.
x4 = x1 + x3
x5 = x2 – x4
Data Analyis
This section will include all data collected during the lab.
|
Thermochemical Data |
||||||
|
|
Tinitial (°C) |
Tfinal (°C) |
ΔT (°C) |
moles NaOH |
qreaction (kJ) |
ΔHrxn |
|
Reaction 1 |
25.0
|
30.3
|
+5.3
|
0.025 |
-1.11 |
-44.4 |
|
Reaction 2 |
25.0
|
37.0
|
+12.0
|
0.025 |
-2.51 |
-100.4 |
|
Reaction 3 |
25.0
|
31.7
|
+6.7
|
0.025 |
-1.40 |
-56.1 |
Reaction 1: NaOH(s) → Na+(aq) + OH-(aq) + x1 kJ
1g /39.977g/mol = 0.025 moles
Moles NaOH = 0.025
qsolution = (4.184 J/g °C) (50.0g) (30.3°C -25.0°C)
= -1108.76 J/ 1000
qreaction (kJ) = -1.11 kJ
ΔH = -1.11 kJ/ 0.025 moles
ΔHrxn = -44.4 kJ/mol
Reaction 2: NaOH(s) + HCl(aq) → H2O(l) + Na+(aq) + Cl-(aq) + x2 kJ
1g /39.977g/mol = 0.025 moles
Moles NaOH = 0.025
qsolution = (4.184 J/g °C) (50.0g) (37.0°C -25.0°C)
= -2510.40 J/ 1000
qreaction (kJ) = -2.51 kJ
ΔH = -2.51 kJ/ 0.025 moles
ΔHrxn = -100.4 kJ/mol
Reaction 3: NaOH(s) + HCl(aq) → H2O(l) + Na+(aq) + Cl-(aq) + x3 kJ
1g /39.977g/mol = 0.025 moles
Moles NaOH = 0.025
qsolution = (4.184 J/g °C) (50.0g) (31.7°C -25.0°C)
= -1401.64 J/ 1000
qreaction (kJ) = -2.51 kJ
ΔH = -1.40 kJ/ 0.025 moles
ΔHrxn = -56 kJ/mol
Net Ionic Equations
Reaction 1: NaOH(s) → Na+(aq) + OH-(aq) -44.4 kJ
Reaction 2: NaOH(s) + HCl(aq) → H2O(l) + Na+(aq) + Cl-(aq) - 100.4 kJ
Reaction 3: NaOH(s) + HCl(aq) → H2O(l) + Na+(aq) + Cl-(aq) -56 kJ
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