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)
Partial pressure ?
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.
Calculate the value of Kp for this dissociation reaction at this temperature.
回答 (1)
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.)
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.)
收錄日期: 2021-05-01 22:45:30
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