Answer to Question #263803 in General Chemistry for Salem

Question

When 10.0 mL 0.100 M NaOH and 5.00 mL 0.200 M HCl were mixed in a coffee-cup calorimeter (heat capacity = 46.5 J °C-1), the temperature increased by 1.20°C. Determine the change in enthalpy of the reaction (in kJ mol-1)

Answer 1

We can use the expression q = Ccal∆T to find the heat of this reaction.

q = heat = ?

Ccal = heat capacity of calorimeter

∆T = change in temperature = 1.20º

q = (46.5J/º)(1.2º)

q = 55.8 J

However, the question is asking for enthalpy of reaction in kJ/mol so we need to find moles and then divide our calculate heat by the number of moles.

mols NaOH = 10.0 ml x 1 L/1000 ml x 0.100 mol/L = 0.001 mols

mols HCl = 5.00 ml x 1 L/1000 ml x 0.200 mol/L = 0.001 mols

∆Hrxn = 55.8 J / 0.001 mols = 55,800 J = 55.8 kJ/mol