Answer:
Explanation:
1) Word equation (given):
• Dinitrogen gas + dihydrogen gas → gaseous ammonia.
2) Chemical equation:
• N₂ (g) + 3H₂ (g) → 2NH₃ (g)
3) Mole ratios:
• 1 mole N₂ (g) : 3 mole H₂ (g) : 2 mole NH₃ (g)
4) Reaction rates:
• Rate of dinitrogen consumption: r₁ = n₁ / t (moles/s)
• Rate of ammonia production: r₂ = n₂ / t (moles/s)
• Due to the stoichiometric ratios: r₂ = 2 × r₁
5) Calculate r₁:
• 284 liter / s
• PV = nRT ⇒ n = PV / (RT)
• Divide by time, t: n/t = P (V/t) / (RT)
• Substitute V/t = 284 liter/s, P = 0.75 atm, and T = 196 +273.15K = 469.15K
r₁ = n₁ / t = (0.75 atm) (284 liter/s) / [ (0.08206 atm-liter/K-mol) (469.15k) ]
= 5.53 moles/s
6) Calcualte r₂
7) Convert rate in mole/s to rate in kg/s