Calculating Enthalpy in a Calorimeter Experiment

Question

When 25.0 mL of 1.0 M H2SO4 is added to 50.0mL of 1.0M of NaOH at 25.0 degrees C in a calorimeter, the temperature of the aqueous solution increases to 33.9 degrees C. Assuming that the specific heat of the solution is 4.18J/(g*degree C), that its density is 1.00g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate enthalpy in kilojoules for that reaction.

 

H2SO4 (aq) + 2NaOH (aq)---> 2H20(l)+ Na2SO4 (aq) 

Answer 1

q = mC∆T

m = mass = 25.0 ml + 50.0 ml = 25.0 g + 50.0 g = 75.0 g since the density is 1 g/ml

C = specific heat = 4.18 J/g/deg

∆T = change in temperature = 33.9 degrees C

q = (75 g)(4.18 J/g/deg)(33.9 deg) = 10628 J = 10.6 kJ heat for this reaction.

If you want enthalpy per mole, then you need to divide this by moles NaOH or H2SO4 present (they are the same), and that would be 0.025 L x 1 mol/L x 2  = 0.005 moles or 0.05 L x 1.0 mole/L = 0.005 (x2 for H2SO4 b/c 2 H+ per mol)