Q50 / 87 · NTA Abhyas (JEE Main) 2020 · Chemistry · Chemical Equilibrium

Hydrogen peroxide (H2O2 ) decomposes according to the equation

2 H 2 O 2 ⇌2 H 2 O I + O 2 g

From the following data at 2 5 ∘ C   calculate the value of Kp at 400 K for the above reaction, Δ H ∘ = - 1 9 6 · 0  KJ    Δ S ∘ = 1 2 5 · 6 5 J/K .
[Given: 10 0.15 =1.41 ]

  • Correct answer
    A
    0.14 x 1032
  • B
    0.14 x 10-32
  • C
    0.14 x 103
  • D
    1.3 x 1015

Solution

Δ G ∘ = + Δ H ∘ - T Δ S ∘

          = - 1 9 6 0 0 0 - 4 0 0 1 2 5 · 6 5

          = - 1 9 6 0 0 0 - 5 0 2 6 0

          = - 2 4 6 2 6 0 J = - 2 · 3 0 3 × 8 · 3 1 4 × 4 0 0  log  K p 

log  K p = - 2 4 6 2 6 0 - 7 6 5 8 = 3 2 · 1 5

      K p = Antilog 3 2 · 1 5

          = 0.14 × 1 0 3 2

Hence, the correct answer is option: 'A'.

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