Question

in a closed rigid vessel initally pure NH3 is presnt at Tk. at a given temp, NH3 decomposes to produse N2 and H2 decompositio folows 1st order kinetics. the rxn is 50% completedin 14 mins. after infinite time the pressue is 2 atm. after how much time will pressure become 1.8 atm

Answer 1

It takes approximately 32.5 minutes for the pressure to become 1.8 atm.

Answer 2

Solution:

For the decomposition of NH₃, the balanced reaction is
  2 NH₃(g) → N₂(g) + 3 H₂(g).

Let the initial moles of NH₃ = 2 (for convenience). If a fraction y of NH₃ decomposes then:

• Moles of NH₃ remaining = 2(1 – y)
• Moles of products formed = N₂: y and H₂: 3y
• Total moles at time t = 2(1 – y) + y + 3y = 2 + 2y

Since the vessel is rigid and isothermal, pressure ∝ number of moles. Let initial pressure P₀ correspond to 2 moles. Then,   P₀ ∝ 2  and  P∞ ∝ 2 + 2(1) = 4.
Given that P∞ = 2 atm, it follows that P₀ = 1 atm and the pressure at time t is

  P(t) = 1 atm × (1 + y).

At 50% completion, y = 0.5, so:   P = 1(1 + 0.5) = 1.5 atm
and the half-life t₁/₂ = 14 min. Because the reaction is first order in NH₃, we have   [NH₃] = [NH₃]₀ e^(–kt).

Since the moles of NH₃ remaining are 2(1 – y), the fraction unreacted is   (1 – y) = e^(–kt).

For the pressure of 1.8 atm at some time t,   1 + y = 1.8  ⟹  y = 0.8.

Thus,
  e^(–kt) = 1 – y = 0.2  ⟹  –kt = ln(0.2)  ⟹  t = –(1/k) ln(0.2) = (ln 5)/k.

Using the half-life relation for first order kinetics,   t₁/₂ = ln 2/k  ⟹  k = (ln 2)/14.

Substitute k into the time expression:   t = (ln 5) / (ln 2/14) = 14 (ln 5)/(ln 2).

Numerically, with ln 5 ≈ 1.609 and ln 2 ≈ 0.693:   t ≈ 14 × (1.609 / 0.693) ≈ 14 × 2.322 ≈ 32.5 min.

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Minimal Explanation:

1. In a rigid vessel, total moles and hence pressure is proportional to (initial moles + moles from reaction) = 2 + 2y.
2. With P₀ = 1 atm and final pressure 2 atm, P(t) = 1(1 + y). At 1.8 atm, y = 0.8.
3. First order decay: (1 – y) = e^(–kt); with t₁/₂ = 14 min, k = ln2/14.
4. Solve: e^(–kt) = 0.2 ⟹ t = 14 ln5/ln2 ≈ 32.5 min.