Question

11. Zinc metal reacts with hydrochloric acid according to the following balanced equation: Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g) When 0.103 g of Zn (s) is combined with enough HCl to make 50.00 mL of solution in a coffee cup calorimeter, all of the zinc reacts raising the temperature of the solution from 22.50 oC to 23.70 oC. Calculate Hrxn in kJ/mol of Zn. (Assume that density of the solution = 1.00 g/mL and specific heat capacity of the solution = 4.184 J/g oC)

Answer 1

Ans. Moles of Zn = Mass / Molar mass

                                    = 0.103 g / (136.2954 g/mol)

                                    = 7.5571 x 10^-4 mol

# Total mass of reaction mixture = 50.0 g (HCl) + 0.103 g

                                    = 50.103 g

# The amount of heat gained by reaction mixture to increase its temperature is given -

            q = m s dT                          

Where,

q = heat

m = mass

s = specific heat

dT = Final temperature – Initial temperature

Putting the values in above reaction-

            q = 50.103 g x (4.184 J g^-10C^-1) x (23.70 – 22.50)⁰C

            Hence, q = 209.630952 J

# The total amount of heat gained by the reaction mixture must be equal to the amount of heat released during reaction of Zn metal with HCl.

So,

Amount of heat released during reaction of Zn with HCl, Hrxn = - 209.630952 J

Note that the –ve sign of Hrxn indicates that heat is being released during reaction of Zn with HCl.

Now,

            Molar Hrxn = Total heat released / Moles of Zn consumed

                                    = (-209.630952 J) / (7.5571 x 10^-4 mol)

                                    = -2.7740 x 10⁵ J/ mol

                                    = -277.40 kJ/mol