Calculate the amount of heat evolved when[1] 500cm3 of 0.1 M HCl acid is mixed with 200cm3 of 0.2 M NaOH solution [2] 200cm3 of 0.2 M H2SO4 is mixed with 400cm3 of 0.5 M KOH solution. Assuming that the specific heat of water is 4.18 K-1 g-1 ignoring the heat absorbed by the container, therometer, stirrer etc., what would be the rise is temperature in each of the above cases?
1) The neutralization reaction between HCl and NaOH can be written as :
HCl (aq) + NaOH (aq)----> NaCl (aq) + H2O (l)...ΔH = -57.1 kj/mole
Number of moles of HCl in given solution = molarity x volume of solution in L
= 0.1 x 500/1000
= 0.05 moles
Number of moles of NaOH in solution = 0.2 x 200/1000
= 0.04 moles.
As the balanced reaction indicates 1 mole of HCl need 1 mole of NaOH for complete neutralization therefore only 0.04 moles of NaOH will neutralize 0.04 moles of HCl.
Amount of heat releasesd when 1 mole of HCl reacts with 1 mole of NaOH = 57.1 KJ
Therefore amount of heat released when 0.04 mole of HCl reacts completely = 57.1 KJ /1 x 0.04 mole
= 2.292 KJ.
Rise in temperatuture
=
c = specific heat of water
m = mass of the mixture
= = 2.292 KJ / 4.18 KJ K-1 kg-1 x 0.7 kg
= 0.783 K
2) The neutralization reaction between H2SO4 and KOH can be written as :
H2SO4 (aq) + 2 KOH (aq)----> K2SO4 (aq) + 2 H2O (l)...ΔH = -57.1 kj/mole
Number of moles of H2SO4 in given solution = molarity x volume of solution in L
= 0.2 x 200/1000
= 0.04 moles
Number of moles of KOH in solution = 0.5 x 400/1000
= 0.2 moles.
As the balanced reaction indicates 1 mole of H2SO4 need 2 mole of KOH for complete neutralization therefore 0.04 moles of H2SO4 will be used in neutralization of 0.08 moles of KOH.
Amount of heat releasesd when 1 mole of H2SO4 reacts with 2 mole of KOH = 57.1 KJ x 2 (because 2 moles of water is formed)
= 114.2 KJ
Therefore amount of heat released when 0.04 mole of H2SO4 would reacts completely = 114.2 KJ /1 x 0.04 mole
= 4.568 KJ.
Rise in temperatuture=
c = specific heat of water
m = mass of the mixture
= = 4.568 KJ / 4.18 KJ K-1 kg-1 x 0.6 kg
= 1.8 K
HCl (aq) + NaOH (aq)----> NaCl (aq) + H2O (l)...ΔH = -57.1 kj/mole
Number of moles of HCl in given solution = molarity x volume of solution in L
= 0.1 x 500/1000
= 0.05 moles
Number of moles of NaOH in solution = 0.2 x 200/1000
= 0.04 moles.
As the balanced reaction indicates 1 mole of HCl need 1 mole of NaOH for complete neutralization therefore only 0.04 moles of NaOH will neutralize 0.04 moles of HCl.
Amount of heat releasesd when 1 mole of HCl reacts with 1 mole of NaOH = 57.1 KJ
Therefore amount of heat released when 0.04 mole of HCl reacts completely = 57.1 KJ /1 x 0.04 mole
= 2.292 KJ.
Rise in temperatuture
c = specific heat of water
m = mass of the mixture
= 0.783 K
2) The neutralization reaction between H2SO4 and KOH can be written as :
H2SO4 (aq) + 2 KOH (aq)----> K2SO4 (aq) + 2 H2O (l)...ΔH = -57.1 kj/mole
Number of moles of H2SO4 in given solution = molarity x volume of solution in L
= 0.2 x 200/1000
= 0.04 moles
Number of moles of KOH in solution = 0.5 x 400/1000
= 0.2 moles.
As the balanced reaction indicates 1 mole of H2SO4 need 2 mole of KOH for complete neutralization therefore 0.04 moles of H2SO4 will be used in neutralization of 0.08 moles of KOH.
Amount of heat releasesd when 1 mole of H2SO4 reacts with 2 mole of KOH = 57.1 KJ x 2 (because 2 moles of water is formed)
= 114.2 KJ
Therefore amount of heat released when 0.04 mole of H2SO4 would reacts completely = 114.2 KJ /1 x 0.04 mole
= 4.568 KJ.
Rise in temperatuture
c = specific heat of water
m = mass of the mixture
= 1.8 K