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If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the
rate of production of nitrogen and hydrogen? Given 2NH3 3H2 + N2

Sagot :

dsdrajlin

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

The Answer is: 3.15 × 10⁻⁶ mol H₂/L.s; 1.05 × 10⁻⁶ mol N₂/L.s

Step 1:  When we Write the balanced equation

  • Then 2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

  • After that, The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
  • Now 2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

  • After that, The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
  • Now 2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

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