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

The standard enthalpy of the reaction is -248.8 kJ/mole

Step-by-step explanation:

Step 1: The balanced equation

Zn(s) + 2HBr(aq) → ZnBr2(aq) + H2(g)

This means for 1 mole of Zinc we need 2 moles HBr consumed, to produce 1 mole of ZnBr2 and 1 mole of H2

Step 2: Calculate number of moles of Zinc

Number of moles of Zinc = mass of Zinc / Molar mass of zinc

Number of moles of Zinc = 2.50 grams / 65.38 g/mol = 0.038 moles

Step 3: Calculate moles of HBr

for 1 mole of Zinc we need 2 moles HBr consumed, to produce 1 mole of ZnBr2 and 1 mole of H2

Number of moles of HBr = Molarity * volume

Number of moles HBr = 2 M * 0.1 L = 0.2 moles

Step 4: find the limiting reactant

Zinc is the limiting reactant, there will react 0.038 moles

HBr is the reactant in excess, there will remain 0.2 -2* 0.038 = 0.124 moles

Step 5: Calculate heat absorbed by calorimeter.

Heat absorption = tot. heat capacity * Temperature

Q = 448 J/K * 21.1 K

Q = 9452.8 J

Step 6: Calculate heat per mole of Zn

ΔH = 9452.8 J /0.038 moles = 248757. 9 J/mole = 248.8 kJ/mole

Since heat is given off , it's negative so -248.8 kJ/mole