A100.00 ml of volume of 0.500 m hcl was mixed with 100.00 ml of 0.500 m koh in a constant pressure calorimeter. the initial temperature of individual acid and base was 23.0 degree celsius. the final temperature after mixing was 25.5 degree celsius. given the specific heat of the solution was 3.97 j. g-1. deg.-1
a) calculate the heat of the reaction in kj.
b) calculate the enthalpy of the reaction in kj/mol.

The enthalpy of the reaction in kJ/mol is 39.7 kJ/mol.

The equation of the reaction is;

HCl(aq) + KOH(aq) > KCl(aq) + H2O(l)

We have the following information from the question;

Final volume of solution = 100.00 ml HCl + 100.00 ml KOH = 200.00 ml

Mass of solution = 200.00 g of solution

Number of moles of solution = 0.500 M × 100.00/1000 L = 0.05 moles

Notice that the reaction is 1:1

Final temperature of solution = 25.5°C

Initial temperature of the acid and base = 23.0°C

Specific heat of the solution = 3.97 J. g-1 °C

Using the formula;

ΔH = mcθ

ΔH = Heat of reaction

m = mass of solution

c = Specific heat of the solution

θ = temperature rise

Substituting values;

ΔH = 200.00 g × 3.97 J. g-1 °C ×  (25.5°C -  23.0°C)

ΔH = 1.985 KJ

Enthalpy of the reaction in kJ/mol =  1.985 KJ/0.05 moles

= 39.7 kJ/mol

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#a. 1.95 kJ

#b. 39 kJ/mol

We are given;

Volume of HCl =100 mL

Molarity of  HCl = 0.5 M

Volume of KOH = 100 mL

Molarity of KOH = 0.5 M

Change in temperature, Δt, = 2.5 °C

Specific heat capacity of the solution = 3.97 J/g°C

We are required to calculate the heat of the reaction and the molar enthalpy of the reaction in kJ/mol

Step-by-step solution;

Total volume of the solution = 200 mL

Taking the density as 1 g/mL

Mass of the solution = 200 mL× 1 g/mL

                                  = 200 g

Heat change = m×c×Δt

Therefore;

Heat change, Q = 200 g × 3.9 J/g°C × 2.5 °C

                          = 1950 Joules

But, 1 kJ = 1000 J

Therefore, heat change = 1.95 kJ

Moles = Molarity × Volume

Moles of HCl = 0.5 × 0.1 L

                     = 0.05 moles

But since moles of the acid are equal to the moles of water produced.

Moles of solution = 0.05 moles

But;

ΔH = heat change ÷ Number of moles

     = 1.950 kJ÷ 0.05 moles

    = 39 kJ/mol

Therefore, the enthalpy of the reaction is 39 kJ/mol