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Question

A 2.50 g sample of powdered zinc is added to 100.0 mL of a 2.00 M aqueous solution of hydrobromic acid in a calorimeter. The total heat capacity of the calorimeter and solution is 448 J/K. The observed increase in temperature is 21.1 K at a constant pressure of one bar. Calculate the standard enthalpy of reaction using these data.

Answer 1

Zn(s) + 2HBr(aq) ==> ZnBr₂(aq) + H₂(g) ... balanced equation for the reaction taking place

Let's find the heat generated:

q = Ccal∆T

q = heat = ?

Ccal = heat capacity of calorimeter = 448 J/K

∆T = change in temperature = 21.2K

q = (448 J/K) (21.2K) = 9453 J

Now that we know the heat generated, we need to find the moles of each compound in the reaction:

mols Zn = 2.50 g Zn x 1 mol Zn / 65.4 g = 0.0382 mols

mols HBr = 100 ml x 1 L/1000 ml x 2.00 mol/L = 0.2 mols

Because the mol ratio in the balanced equation is 1 mol Zn : 2 mol HBr, Zn will be limiting

Since Zn is limiting, the maximum amount of ZnCl2 that can form is 0.0382 mols as seen by ...

0.0382 mols Zn x 1 mol ZnCl2 / 1 mol Zn = 0.0382 mols ZnCl2

To compute the STANDARD ENTHALPY of the reaction, we must use the basis of ONE MOLE since that is part of the definition of "standard" enthalpy.

heat generated = 9453 J

moles = 0.0382

Standard ∆Hrxn = 9453 J / 0.0382 mols = 247,461 J/mol = 247 kJ/mol