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

Solution

Given data

Moles of H2= 30

Moles of O2 = 15

Pressure = 101.3 kPa = 1 atm

Initial temperature = 20 C

Lets calculate the moles of the H2O(g) produced by the reaction using the mole ratio of the both reactants.

Balanced reaction equation is as follows

2H2 +O2 ----> 2H2O + 241.8kJ

Calculating using moles of H2

(30 mol H2 * 2 mol H2O) / 2 mol H2 = 30 mol H2O

Calculating using moles of O2

( 15 mol O2 * 2 mol H2O)/ 1 mol O2 = 30 mol H2O

Both reactant produces the same moles of H2O

So the moles of H2O produced = 30 moles

Now lets calculate the amount of energy produced by the reaction

2 mol H2O = 241.8 kJ

30 mol H2O= ? kJ

(30 mol H2O * 241.8 kJ) / 2 mol H2O = 3627 kJ

Lets convert kJ to Joules

(3627 kJ * 1000 J) / 1 kJ = 3627000 J

Now lets calculate amount of heat remaining in the container to heat the water vapor

90 % is lost by the container walls

Therefore only 10 % is remaining for the heating water vapor

So ,

(3627000 J * 10 %) / 100 % = 362700 J

So amount of heat remaining to heat the water vapor is 362700 J

Now lets calculate mass of the water vapor

Mass in gram = moles * molar mass

Mass of water vapor = 30 mol * 18.0148 g mol-1

Mass of water = 540.4 g

Now lets use the mass of water vapor and the specific heat of water vapor to calculate the final temperature of the water vapor

Formula is as follows

q=m*c s * (Tf