Question

Ammonia is produced industrially from the reaction of hydrogen with nitrogen under pressure in a sealed reactor. What is the percent decrease in pressure of a sealed reaction vessel during the reaction between 3.60 x 10^3 moles of H2 and 1.20 x 10^3 moles of N2 if half of the N2 is consumed? The correct answer is 25%.

Answer 1

Solution : -

Balanced reaction equation

N2 + 3H2 ----- > 2NH3

Moles of N2 = 1.20*10^3

Moles of H2 = 3.60*10^3

Solets calculate the moles of the N2 needed to react with given moles of H2

3.60 mol H2 * 1 mol N2 / 3 mol H2 = 1.20*10^3 mol N2

So all of the N2 is consumed

Now lets calculate the moles of the NH3 that can be formed

3.60*10^3 mol H2 * 2 mol NH3 / 3 mol H2 = 2.40*10^3 mol NH3

Total moles of reactant present initialy = 3.6*10^3 + 1.2*10^3 = 4.8*10^3 moles

When the half of the N2 is reacted then

Moles of the product formed are = (1.2*10^3 mol N2 /2)*(2 mol NH3 / 1 mol N2) = 1.2*10^3 mol NH3

Moles of H2 reacted are = (1.2*10^3 mol N2 /2)*(3 mol H2 / 1 mol N2) = 1.8*10^3 mol H2

So now after ther reaction

Moles of H2 remain = 3.6*10^3 – 1.8*10^3 = 1.8*10^3

Moles of N2 remain = 1.2*10^3 / 2 = 6.0*10^2

Moles of NH3 formed = 1.2*10^3 mol

So total moles after reaction = 1.8*10^3 + 6.0*10^2 + 1.2*10^3 = 3.6*10^3 mol

So the change in the total moles of the reaction mixture = 4.8*10^3 mol – 3.6*10^3 mol = 1.2*10^3 mol

So the change in the percent of the moles =( 1.2*10^3 mol / 4.8*10^3 mol )*100%

                                                                   = 25 %

Since pressure is directly proportional to the moles of the gas therefore pressure will also decrease by 25%