chemistry/thermodynamics/calorimetry

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/ºC . The observed increase in temperature is 21.1ºC at a constant pressure of 1 atm. Using these data, calculate the standard enthalpy of reaction in kJ/mol. 

Zn (s) + 2 HBr (aq) → ZnBr₂ (aq) + H₂ (g)

Answer 1

Zn (s) + 2 HBr (aq) → ZnBr₂ (aq) + H₂ (g) ... balanced equation

Whenever you are given the mass or moles of BOTH reactants, it becomes imperative to find the limiting reactant, if any. One easy way to do this is to divide the moles of reactant by its coefficient, as follows:

Zn: 2.50 g x 1 mol/65.4 g = 0.0382 moles Zn (÷1=0.0382)

HCl: 0.100 L x 2.00 mol/L = 0.200 moles HCl (÷2=0.100)

Zn would be limiting under these conditions. Now we can go ahead and use the moles of Zn to answer the question.

heat = q = 448 J/º x 21.1º = 9453 J = heat of reaction

To get this on a per mol Zn basis, we simply divide by the moles of Zn that are present:

Standard enthalpy of reaction = 9453 J / 0.0382 mol Zn = 247,461 J/mol = 247 kJ/mole