Question

The reaction of an acid such as HCl with a base, such as NaOH, in water...

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

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Answer #1

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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