Solving Example 7.6 p.273
th
Jespersen 6 Edition
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Example 7.6 p.273
HCl(aq) + NaOH(aq)
H2O(l) + NaCl(aq)
A student placed 50.0 mL of 1.00 M HCl at 25.5oC in
a coffee cup calorimeter. To this was added 50.0
mL of 1.00 M NaOH soln also at 25.5oC. The
mixture was stirred and temp quickly increased to a
max of 32.2oC. What is ΔH in kJ per mole of HCl?
Because the solns are relatively dilute, we can
assume that their specific heats are close to that of
water, 4.18 J g-1 oC-1. The density of 1.00 M HCl is
1.02 g/mL and that of 1.00 M NaOH is 1.04 g/mL.
(We will neglect the heat lost to the Styrofoam
itself, to the thermometer or to the surrounding air.)
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HCl(aq) + NaOH(aq)
H2O(l) + NaCl(aq)
50.0 mL 50.0 mL
1.00 M
1.00 M
25.5oC
25.5oC
32.3oC
s of solutions = 4.18 J/(goC)
D=1.02g/mL 1.04 g/mL
Question with 2 systems, so
q
= –q
qrxn = – (qsoln + qcalorim)
for coffee cup calorimeter qcalorim = 0 J
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(Styrofoam cups do not absorb much heat.)
qrxn = – (qsoln + qcalorim)
qrxn = – (qsoln + 0)
qsoln = s m T where s = 4.18 J/(g.oC)
m= mass of soln = ?
T = (32.2-25.5)oC = 6.7oC
Mass of soln = massHCl + massNaOH
m of HCl = 50.0 mL HCl x 1.02 g/mL = 51.0 g
m of NaOH = 50.0 mL NaOH x 1.04 g/mL = 52.0 g
m of soln = 51.0 g + 52.0 g = 103.0 g
qsoln = s m T = (4.18 J/(g.oC))(103.0g)(6.7oC)
= 2,884.6 J = 2.88 kJ
Ques asked for H in kJ per mole HCl. Is this for 1 mol HCl?
NO! Our ans of 2.88 kJ is for how many moles?
x mol HCl = 0.0500 L x 1.00 mol HCl/L = 0.0500 mol HCl
 2.88 kJ 
x kJ  1 mol HCl 
 = 57.6 kJ
 0.0500 mol 
Is that H ? ... Not quite...
q rxn = - q soln = - 58 kJ/mol HCl Ans. H = - 58 kJ/mol HCl
For an open vessel, q = H
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