Question

Into a 40.0 L closed container held at 155°C were introduced 5.60 g of gaseous nitrogen and 1.41 g of gaseous hydrogen. A spark was used to initiate a reaction between hydrogen and nitrogen; gaseous ammonia was the sole product of the reaction. What was the pressure inside the container at the conclusion of the reaction, assuming that the reaction occurred to the maximum possible extent? Assume each substance behaved ideally.

Answer 1

Mass of nitrogen = 5.60 grams

moles of nitrogen = mass/MW = 5.6/28 = 0.2

Mass of hydrogen = 1.41 grams

moles of hydrogen = mass/MW = 1.41/2 = 0.705

N₂ (g)+ 3 H₂ (g) ---------> 2 NH₃(g)

From the given data nitrogen is the limiting reagent

1 mole of N₂ reacts with 3 moles of H₂ to give 2 moles of NH₃

0.2 mole of N₂ reacts with 0.6 moles of H₂ to give 0.4moles of NH₃

moles of unreacted hydrogen = 0.705-0.6 =0.105

Total number of moles in the container n= 0.2+0.6+0.4+0.105 =1.305

T = 155°C = 155+273 = 428 K V=40.0 L   R = 0.0821 L.atm.K^-1.mol^-1.

PV = nRT

P=nRT/V = 1.305 x 0.0821 x 428/40 = 1.1464 atm.

Pressure inside the container at the conclusion of the reaction = 1.1464 atm.