Partial pressure ?

👨🏻‍🏫 學術台化學

[匿名]

2020-09-15 12:59 pm

A sample of is placed in an empty cylinder at a certain temperature. After equilibrium is reached the total pressure is 2.4 atm and 15% (by moles) of the original has dissociated to .

Calculate the value of Kp for this dissociation reaction at this temperature.

更新1:

A sample of N2O4(g) is placed in empty cylinder... ... the original N2O4(g) has dissociated to NO2(g)

回答 (1)

胡雪8°

2020-09-15 4:46 pm

              N₂O₄(g) ⇌ 2NO₂(g) Kp
Initial:      1 mol     0 mol
Change: -0.15 mol   +0.30 mol
At eqm:   0.85 mol      0.30 mol

At eqm:
Partial pressure of N₂O₄, 𝑃N₂O₄ = 𝑃total • 𝑋N₂O₄ = (2.4 atm) • [0.85/(0.85+0.3)] = 1.77 atm
Partial pressure of NO₂, 𝑃NO₂ = 𝑃total•𝑋NO₂ = (2.4 atm)•[0.30/(0.85+0.30)] = 0.626 atm

Kp = (𝑃NO₂)²/ 𝑃N₂O₄ = 0.626²/1.77 = 0.22 (to 2 sig. fig.)

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