When 200 g of NaOH is added to 500 ml of 10 M HCl solution i

When 2.00 g of NaOH is added to 50.0 ml of 1.0 M HCl solution in a constant pressure calorimeter, the temperature increased by 13.1^oC. Calculate the heat of neutralization in KJ/mol. (S.H.C. for the solution is 4.18 J/K.g, assume the calorimeter does not absorb any heat and the density of the solution is 1.0 g/mL).

Solution

Moles of acid HCl = 0.05 mol

Moles of NaOH = 2/40 = 0.05 mol

thus acid is completely neutralised .

m = 50 mL solution = 50 g, (d = 1g/ml)

delta T = rise in temperature = 13.1 degree C , given

S.H.C = 4.18 J/K g

so, heat = m* S.H.C*delta T= 50*4.18*13.1 = 2737.9 J

hence, heat evolved in the neutralisation of 1 mole acid

= 2737.9/0.05 = 54758 J/mol = 54.758 KJ/mol