Given the following data: Solution Conc. (M) Volume (mL) Tinitial (ºC) Tfinal (ºC) NaOH 0.750 50.00...
Given the following data:
|
Solution |
Conc. (M) |
Volume (mL) |
Tinitial (ºC) |
Tfinal (ºC) |
|
NaOH |
0.750 |
50.00 |
19.84 |
23.18 |
|
HCl |
0.500 |
50.00 |
19.84 |
Calculate ∆Hº for the reaction:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
You may assume all solutions to have the same density and specific heat as water (1.00 g/mL and 4.184 J/gºC)
Answer in kJ/mol
Homework Answers
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
total volume = 50 +50 = 100ml
mass of solution = volume * density
= 100*1 = 100g
q = mc
T
= 100*4.184*(23.18-19.84) = 1397.45J
∆Hº = 1397.45J
no of moles of HCl = molarity * volume in L
= 0.5*0.05 = 0.025 moles
no of moles of NaOH = molarity * volume in L
= 0.75*0.05 = 0.0375mole
HCl is limiting reactant
the reaction enthalpy per mole = 1397.45/0.025 = 55898J/mole = 55.898Kj/mole
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Given the following data:
Solution
Conc. (M)
Volume (mL)
Tinitial (ºC)
Tfinal (ºC)
NaOH
0.750
50.00...
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