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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

Hydrogen peroxide (H_ 2 O_ 2 ) decomposes according to the equation...

Hydrogen peroxide (H_{2}O_{2} ) decomposes according to the equation 2H2O2⇌2H2O(I)+O2(g)Fromthefollowingdataat25∘CcalculatethevalueofKpat400Kfortheabovereaction,Δ2{H}_{2} {O}_{2} ⇌ 2{H}_{2} \text{O} \left(\text{I}\right) + {O}_{2}\left(\text{g}\right) From the following data at 2 {5}^{∘} \text{C} calculate the value of K_{p} at 400 K for the above reaction, \Delta H^{∘}=- 1 9 6 ⋅0KJΔ\cdot0 \text{KJ} \Delta S^{∘}=1 2 5 ⋅65J/K.[Given:\cdot6 5 \text{J/K}. [Given:{10}^{0.15}=1.41]]

Options

  1. A0.14 x 10^{32}
  2. B0.14 x 10^{-32}
  3. C0.14 x 10^{3}
  4. D1.3 x 10^{15}

Correct answer

A. 0.14 x 10^{32}

Step-by-step solution

ΔG∘=+ΔH∘−TΔS∘\Delta G^{∘} = + \Delta H^{∘} - \text{T} \Delta S^{∘}

=−196000−400(125⋅65)= - 1 9 6 0 0 0 - 4 0 0 \left(1 2 5 \cdot6 5\right)

=−196000−50260= - 1 9 6 0 0 0 - 5 0 2 6 0

& & = - 2 4 6 2 6 0 \text{J} = -2 \cdot3 0 3 \times8 \cdot3 1 4 \times4 0 0 \text\log {& K}_{p& }

\text\log {K}_{p} = \dfrac{- 2 4 6 2 6 0}{- 7 6 5 8} =3 2 \cdot1 5

Kp=Antilog(32⋅15){K}_{p} = \text{Antilog} \left(3 2 \cdot1 5\right)

=0.14×1032= 0.14 \times1 {0}^{3 2}
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