In: Chemistry
When 40.0 mL of 0.580 M H2SO4 is added to 40.0 mL of 1.16 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:
Number of moles of H2SO4 is , n = molarity x volume in L
= 0.580 M x (40.0/1000 ) L
= 0.0232 mol
Number of moles of KOH is , n' = Molarity x volume in L
= 1.16 M x (40.0 / 1000 ) L
= 0.0464 mol
H2SO4 + 2 KOH K2SO4 + 2H2O
According to balanced chemical reaction,
1 mole of H2SO4 reacts with 2 moles of KOH
0-0232 mole of H2SO4 reacts with 2x0.0232=0.0464 moles of KOH
So all the masses of H2SO4 &KOH completly reacted.
Molar mass of H2SO4 is = (2xAt.mass of H )+ At.mass of S + (4xAt.mass of O)
= (2x1) + 32 + (4x16)
= 98 g/mol
So mass of H2SO4 reacted , m = number of moles x molar mass
= 0.0232 mol x 98 (g/mol)
= 2.27 g
The amount of heat evolved is , Q = mcdt
Where
m = mass of H2SO4 reacted = 2.27 g
c = specific heat capacity of the solution = c = 4.184 J/g×K
dt = change in temperature
= final temeparture - initial temperature
= 30.17 oC - 23.50 oC
= 6.67 oC
Plug the values we get Q = 2.27 g x 4.184 J/g×K x 6.67 oC
= 63.3 J
So the heat evolved is 63.3 J of heat when 2.27 g of H2SO4 reacted
So the amount of heat evolved for 98 g of H2SO4 is M
M = (63.3 x 98 ) / 2.27
= 2734.9 J
So the enthalpy change of H2SO4 is 2734.9 J/(98g)
= 2734.9 J/mol