2.00 moles of NO2 are placed in a 1.00 L flask and allowed to react. At equilibrium 1.80 moles NO2 are present. Calculate the Keq
Pls help
Given equilibrium: 2NO2(g) <-> N2O4(g)?
2019-04-29 8:50 pm
回答 (1)
2019-04-29 9:05 pm
____________________ 2NO₂(g) ____ ⇌ ____ N₂O₄(g) ____ Keq
Initial (mol/L): _________ 2.00 _______________ 0
Change (mol/L): ________ -2y _______________ +y
Equilibrium (mol/L): __ 2.00 - 2y ______________ y
At equilibrium, moles of NO₂ = (2.00 - 2y) mol = 1.80 mol
y = 0.1
Keq = [N₂O₄]/[NO₂] = 0.1 / (2.00 - 0.1)² = 0.0309
Initial (mol/L): _________ 2.00 _______________ 0
Change (mol/L): ________ -2y _______________ +y
Equilibrium (mol/L): __ 2.00 - 2y ______________ y
At equilibrium, moles of NO₂ = (2.00 - 2y) mol = 1.80 mol
y = 0.1
Keq = [N₂O₄]/[NO₂] = 0.1 / (2.00 - 0.1)² = 0.0309
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