Calculate the heat change per mole of reaction

Question

React 100.0 mL of 0.400 M hydrochloric acid with 90.0 mL of 0.500 M sodium hydroxide in a bomb calorimeter with a heat capacity of 6.25 kJ/°C. The reaction made the temperature of the thermometer increase by 4.52oC. Calculate the heat change per mole of reaction

Answer 1

This is similar to your previous question. Again, a bomb calorimeter is at constant volume, not constant pressure, so ∆H may not be totally appropriate, since it is at q at constant pressure. Anyway...

Write the correctly balanced equation:

HCl + NaOH ==> NaCl + H₂O .. balanced equation

Find moles of each reactant:

moles HCl = 100.0 ml x 0.400 mols / 1000 mls = 0.0400 mols HCl

moles NaOH = 90.0 ml x 0.500 mols / 1000 mls = 0.045 mols NaOH

HCl is limiting and so 0.0400 mols H₂O will be formed

q = heat = C_cal∆T

q = ?

C_cal = calorimeter constant = 6.25 kJ/º

∆T = change in temperature = 4.52º

Solving for q we have...

q = (6.26 kJ/º)(4.52º) = 28.29 kJ

Molar heat of reaction = 28.29 kJ / 0.0400 mols = -707 kJ / mole (negative since rxn is exothermic)