Question

30 mols of H2 and 15 Mols of O2 at 20oC are placed in a container at 101.3kPa and react exothermically (okay there is a large explosion but ignoring that) if the walls of the container remove 90% of the energy. The equation for the reaction is 2H2 +O2 ? 2H2O + 241.8kJ, what is the temperature of the water vapour produced (c s =2.01J/gC)

Answer 1

2H₂ +O₂ 2H₂O + 241.8kJ

According to the above balanced equation ,

2 moles of H₂ reacts with 1 mole of O₂ produces 241.8 kJ of energy

2x15=30 moles of H₂ reacts with 1x15=15 mole of O₂ produces 241.8x15 = 3627 kJ of energy

But he walls of the container remove 90% of the energy.So 10 % of the energy only is absorbed by the water vapour.

10% of 3627 kJ of energy = 362.7 kJ

So this much of heat is utilized by water vapour.

The heat content , Q = mcdt

Where

m = mass of water vapour = number of moles x molar mass of H₂O

[ NOTE: According to the above balanced equation,2 moles of H₂ produces 2 moles of H₂O

                                                                          30 moles of H₂ produces 30 moles of H₂O ]

m = 30 mol x 18 g/mol

   = 540 g

c = c_s = specific heat capacity = 2.01 J/g^oC

dt = change in temperature

    = final temperature - initial temperature

   = t - 20

Plug the values we get 362.7 kJ = 540 x 2.01 x (t-20)

                           362.7x1000 J = 1085.4t - 21708

                                            t = 354.2 ^oC

Therefore the final temperature of the water vapour is 354.2 ^oC