posted by jonathan
A 5.0g KBr sample, at 25 degrees celsius, dissolved in 25 ml of water, also at 25 degrees celsius. the final equilibrium temperature of the resulting solution is 18.1 degrees celsius, What is the enthalpy change in J/g and KJ/mole of KBr.
q = [mass H2O x specific heat H2O x (Tfinal-Tinitial)]
Substitute and solve for q = delta H for the solution.
q/5 = delta H/g in joules.
delta (H/g) x molar mass KBr = J/mol.
Convert to kJ/mol.