THERD30ak Heat Neutraluation Pre-Laboratory Assignment 1. Read an authoritative
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Question
THERD30ak Heat Neutraluation Pre-Laboratory Assignment 1. Read an authoritative source for a discussion of the safe and proper way to insert a thermometer or glass tubing into a rubber stopper. Read an authoritative source for a discussion of graphing 2. A student doing this experiment was puzzled by the fact that while each of the individual solutions being used is dangerous, the mixture made by combining the two solutions is much less dangerous. Actually, the resulting mixture could be safely disposed of by pouring it into the drain and diluting with a large amount of running water. Describe hazards of each of the why the mixture is not as hazardous as the original solutions from a point of view. 3. A student determined the AHneuran of nitric acid GHNO) mixed with NaoH using the procedure described in this experiment. Fifty milliliters of 1.00M HNO, was added to 505 mL of 1.00M NaoH solution, and the following time-temperature data were recorded. temp, C time, min time, min temp, c 110 31.1 1.0 13.0 31.00 24.9 140 309 308 31.2 5.0 17.0 305 18.0 70 30A 19.0 8.0 31.2 9.0 31.2 100 Make a plot of these time-temperature data.Explanation / Answer
Pre laboratory assignments:
2. It's assumed the mixing reagents are HNO3 and NaOH. Either of them are dangerous under concentrated or dilute conditions but after mixing them we get neutral salt solution (NaNO3) which is not dangerous in solution and can be easily disposed off in a drain under running water. When mixing these reagents one should be very careful as their reaction is exothermic and releases large amount of heat which raises the temerature of the solution.
HNO3 (aq) + NaOH (aq) --------------------> NaNO3 (aq) + H2O (l)
The standard enthalpy (heat) change of neutralization for a strong acid and base = - 57.62 kJ/mol.
Answer for 3.1
Heat of neutralization plot (temerature vs volume of mixture)
Plot could not be uploaded as image uploading window didn't open.
Answer for 3.2.
delta T = Highest Temp Recorded - Initial Room Temp
= 31.5 - 25.0
= 6.5 o C
3.3 Total volume of reaction mixture is total volume of HNO3 and NaOH solutions .
Volume of HNO3 = 50.0 ml
Volume of NaOH = 50.5 ml
Total volume of reaction mixture = 100.5
Answer for 3.4
Mass (m) of reactoin mixture; m = v.d where v = total volume of reaction mixture and d = density of reaction mixture
v = 100.5 ml; d = 1.03 g/ml
Mass (m) of reactoin mixture, m = 100.5 x 1.03 = 103.515 g
Answer for 3.5
The heat (Q) released or transferred during a reaction is
Q = mcp delta T where
m = the mass of the solution = 103.515 g
cp = specific heat capacity of the solution = 3.891 J g-1 deg-1
delta T = the temperature change observed during the reaction = 6.5 o C
Q = 103.515 x 3.891 x 6.5
= 2618.0496 J
heat released or transferred, Q = 2.6180 kJ
Answer for 3.6
Number of moles of HNO3 reacted.
HNO3 solution concentration = 1.0 M
= 1 millimole/mL
HNO3 solution used = 50 mL
Total Number of milli moles of HNO3 used = 50 mL x 1 = 50 milli moles
Similarly Total Number of milli moles of NaOH used = 50.5 ml x 1 = 50.5 milli moles
Since total number of milli moles of HNO3 (50) is less than that used for NaOH, the acid completely neutralized in the reaction. That means number of milli moles of HNO3 reacted during the reaction is 50 milli moles which is equal to 0.050 moles.
Answer for 3.7
delta H = - Heat (Q) transferred during reaction (J) / number of moles of acid reacted
Accordingly Heat (Q) transferred during reaction = 2618.0496 J
number of moles of acid reacted = 0.050 Moles
delta H = - 2618.0496 J / 0.050 Moles
= - 52360.992 J/mole
delta H = - 52.3609 kJ/mole
Answer for 3.8
Percent error = (accepted delta H - calculated delta H) / accepted delta H x 100
accepted delta H of reaction = 58.5 kJ/mole
calculated delta H of reaction = 52.4 kJ/mole
Percent error = 58.5 - 52.4 / 58.5 x 100 = 10.43 %
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