Reaction 1: Sodium hydroxide + Hydrochloric acid |
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Volume (mL) of 2.0 M HCl(aq) used |
25.2 |
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Initial temperature (°C) of the 2.0 M HCl(aq) |
21.63 |
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Volume (mL) of 2.0 M NaOH(aq) used |
24.8 |
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Maximum temperature (°C) of the 2.0 M NaOH(aq) |
21.96 |
NaOH and HCl final temperature: 35.14 degrees C
Determine the Energy (J) absorbed (+) or released (-) by the solution (qsoln). |
Determine the Energy (J) absorbed (+) or released (-) by the calorimeter (qcal). |
Determine the Energy (J) absorbed (+) or released (-) by the reaction (qrxn) |
Calculate the Enthalpy change (kJ/mole), ΔH, for Reaction 1.
Been stuck on these for hours for the same reaction, please help.
The general reaction between NaOH and HCl is
NaOH + HCl -----------> NaCl + H2O
That means 1 mole of NaOH reacts with 1 mole of HCL
Here concentration of HCl taken = 2.0 M
Volume of HCl used = 25.2 mL
Thus mols of HCl taken = conc * volume
= 2.0 M * 25.2 mL
= 2.0 mol/ 1000 ml * 25.2 mL
= 0.0504 mols
Similarly
concentration of NaOH taken = 2.0 M
Volume of NaOH used = 24.8 mL
Thus mols of NaOH taken = conc * volume
= 2.0 M * 24.8 mL
= 2.0 mol/ 1000 ml * 24.8 mL
= 0.0496 mols
That means here NaOH is present in less amount than HCl . Thus it is the limiting reagant. So the reaction will be
0.0496 NaOH + 0.0496 HCl -----------> 0.0496 NaCl + 0.0496 H2O
In this reaction the amount of hear released or absorbed (q) is calculated by the formula-
q = m c ∆T where m = mass of the solution c= specific heat of water = 4.184 J/g°C ∆T = change in temperature
Here m = mass of HCl + mass of NaOH
Now since the final product is water because the product NaCl ldissolves in water, the density of the final solution is taken to be 1g /ml
Thus we can take the density of HCl = density of NaOH = density of solution = 1g/ml
Thus mass of HCl = density * volume = 1g/ml * 25.2 ml = 25.2 g
mass of NaOH = density * volume = 1g/ml * 24.8 ml = 24.8 g
Thus total mass of solution = 25.2 g + 24.8 g = 50 g
The temp change = ∆T = Final temp - initial temp
Here the initial temp of NaOH is taken since after addition of HCl to NaOH, the final temp f the solution increases. Thus here ∆T = Final temp - initial temp
= 35.14 oC - 21.96 oC
= 13.18oC
Thus the heat absorbed by the solution (since temp increases) is =
q = m c ∆T = 50 g * 4.184 J/g°C * 13.18oC
= 2757.25 J
= 2.757 kJ
Thus heat released by the reaction will be opposite of amount of heat absorbed by the solution = - 2.757 kJ
Since here the heat released is -ve, that means the reaction is Exothermic.
Now the Enthalpy change per mole of reaction = - 2.757 kJ / 0.0496 mole
= -55.58 kJ/mole
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