Question

When 24.0 mL of 0.600 M H2SO4 is added to 24.0 mL of 1.20 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate H of this reaction per mole of H2SO4 and KOH reacted. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water: d = 1.00 g/mL and c = 4.184 J/g×K.) H per mole of H2SO4 reacted:

___ kJ/mol

H per mole of KOH reacted:

___kJ/mol

Answer 1

heat released(q) = m*s*DT

m = mass of reaction mixer = 24+24 = 48 g ( density = 1 g/ml)

s = specific heat capacity of the solution = 4.184 j/g.k

DT = 30.17 - 23.5 = 6.67

q = 48*4.184*6.67

   = 1339.55 joule

   = 1.34 kj

q = - DHrxn = 1.34 kj

DHrxn = -1.34 kj

H2SO4(aq) + 2KOH(aq) ----> K2SO4(aq) + 2 H2O(l)

no of mole of H2SO4 = 24/1000*0.6 = 0.0144 mole

no of mole of KOH = 24/1000*1.2 = 0.0288 mole

Enthalpy per mole of H2SO4 = DHrxn / nH2SO4

                      = 1.34/0.0144

                    DH =   - 93 kj/mol

Enthalpy per mole of KOH = 1.34/0.0288

                DH = - 46.53 kj/mol