1. Tardigrade
  2. Question
  3. Chemistry
  4. Hydrogen peroxide (H2O2 ) decomposes according to the equation 2 mathrmH2 mathrmO2 leftharpoons 2 mathrmH2 mathrmO( mathrmI)+ mathrmO2( mathrm~g) From the following data at 2 5° textC calculate the value of Kp at 400 K for the above reaction, Δ textH° = - 1 9 6 ⋅ 0 text KJ Δ textS° = 1 2 5 ⋅ 6 5 textJ/K . [Given: 100.15 = 1.41 ]

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Q. Hydrogen peroxide (H2O2 ) decomposes according to the equation

$2 \mathrm{H}_2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}(\mathrm{I})+\mathrm{O}_2(\mathrm{~g})$

From the following data at $2 5^{^\circ } \text{C}$ calculate the value of Kp at 400 K for the above reaction, $\Delta \text{H}^{^\circ } = - 1 9 6 \cdot 0 \text{ KJ } \Delta \text{S}^{^\circ } = 1 2 5 \cdot 6 5 \text{J/K}$ .

[Given: $10^{0.15} = 1.41$ ]

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

$\Delta \text{G}^{^\circ } = + \Delta \text{H}^{^\circ } - \text{T} \Delta \text{S}^{^\circ }$

$= - 1 9 6 0 0 0 - 4 0 0 \left(1 2 5 \cdot 6 5\right)$

$= - 1 9 6 0 0 0 - 5 0 2 6 0$

$= - 2 4 6 2 6 0 \text{J} = - 2 \cdot 3 0 3 \times 8 \cdot 3 1 4 \times 4 0 0 \text{ log} \text{ K}_{\text{p }}$

$\text{log } \text{K}_{\text{p}} = \frac{- 2 4 6 2 6 0}{- 7 6 5 8} = 3 2 \cdot 1 5$

$\mathrm{K}_{\mathrm{p}}=\operatorname{Antilog}(32 \cdot 15)$

$= 0.1 4 \times 1 0^{3 2}$