Question

Q4. A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The temperature of both solutions before mixing was 23.00°C, and it rises to 26.00°C after the acid-base reaction. What is the enthalpy change for the reaction per mole of salt formed? Assume the densities of the solutions are all 1.08 g/mL and the specific heat capacities of the solutions are 4.18 J/gK. Use the correct sign.

Answer 1

Ca(oH)2(aq) + 2HBr(aq) ----> CaBr2(aq) + 2H2O(l)

1 mole Ca(oH)2 = 2 mol HBr = 1 mol CaBr2

No of mol of Ca(oH)2 = V*M = 50*0.6 = 30 mmol

No of mol of HBr = M*V = 50*0.6 = 30 mmol

LIMITING REACTANT = HBr

No of mol of CaBr2 = 30*1/2 = 15 mmol

heat released in the reaction(q) = m*s*DT

m = mass of solution = (50+50)*1.08 = 108 g

s = specific heat of solution = 4.18 j/g.k

DT = 26-23 = 3

q = 108*4.18*3

    = 1354.32 joule(or) 1.354 kj

DHrxn = -q/n

       = -1.354/(15*10^-3)

       = -90.267 kj/mol