The reaction of an acid such as HCl with a base, such as NaOH, in water is an exothermic reaction.
HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O
In one experiment, a student placed 55.0 mL of 1.00 M HCl in a
coffee-cup calorimeter and carefully
measured its temperature to be 35.5oC. To this was added 55.0 mL of
1.00 M NaOH solution whose
temperature was also 35.5oC. The mixture was quickly stirred, and
the student noticed that the
temperature of the mixture rose to 44.4oC. What was the heat of
reaction (in J/mol)? (Must show all
calculation steps.)
Assumptions for this problem:
1. The system is not perfect. The heat capacity of the calorimeter
is 62.558 J/°C.
2. The molar mass (HCl) = 36.46 g/mol. The molar mass (NaOH) =
39.997 g/mol.
3. Assume that these solutions are close enough to being like water
that their specific heat
capacities are 4.184 J/goC.
4. Assume that the densities of these solutions are 1.00 g/mL
Answer – We are given,
[HCl] = 1.0 M
Volume of HCl = 55.0 mL
[NaOH] = 1.0 M
Volume of NaOH = 55.0 mL
ti = 35.5 oC , tf = 44.4 oC.
density of water = 1.0 g/mL
The specific heat capacity = 4.184 J/goC
Step 1) The Mass of Solution:
Total volume = 55.0 mL+ 55.0 mL = 110 mL
We know the formula for determining the mass from given volume and density,
Step 2) The Heat of reaction
Reaction - HCl + NaOH -----> NaCl + H2O
We know the heat formula
The neutralization reactions are exothermic reaction, hence
∆Horxn = -q
= -4096.1 J
Now we need to calculate the ∆Horxn per moles
Both HCl and NaOH give the same concentration and volume and the mole ratio is 1:1, hence the limiting reactant is both.
Moles of HCl = Moles NaOH = 1.0 M x 0.055 L = 0.055 moles
Thus, the heat of reaction is -74475 J/mol.
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