general chem ch.6

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.

Zn(s)+2HBr(aq)⟶ZnBr2(aq)+H2(g)

Answer 1

heat = ∆T x heat capacity = 21.1K x 448 J/K = 9453 J

This is the amount of heat generated from 2.50 g of Zn and 0.2 moles of HBr (100 ml of 2 M = 0.2 moles)

Standard enthalpies are for 1 mole of substance, so we will use the balanced equation to adjust for these amounts.

moles Zn present = 2.50 g Zn x 1 mol Zn / 65.38 g = 0.03824 mols Zn

moles HBr present = 0.100 L x 2.00 mol/L = 0.200 mols HBr

Limiting reactant is Zn since the balanced equation shows us that 1 mol Zn reacts with 2 mols HBr

So, the reaction taking place would actually be ....

0.03824 mols Zn reacts with 0.07648 mols HBr to produce 0.03824 mols ZnBr2, etc.

Standard enthalpy of reaction = 9453 J / 0.03824 mols = 247,202 J/mol = 247 kJ/mol