edexcel chem u4 questions Watch

mushriqite
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This question is about the reversible reaction below.
N2O4(g) ⇌ 2NO2(g)
(a) A chemist investigating this reaction started with 5 mol of N2O4 and allowed the system to reach equilibrium. If 2 mol of NO2 forms, the amount of N2O4 at equilibrium is (1)
A 1 mol
B 1.5 mol
C 3 mol
D 4 mol
b) Under different conditions, 25% of the moles of gas present at equilibrium is N2O4.If the total pressure of the system is 3 atm, the numerical value of the equilibrium constant Kp is (1)
A 9.00
B 6.75
C 3.00
D 0.15
in b part of the question how the answer is B? plz help
and this question is from edexcel chem unit 4 june 2015 ial question 6
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charco
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(Original post by mushriqite)
This question is about the reversible reaction below.
N2O4(g) ⇌ 2NO2(g)
(a) A chemist investigating this reaction started with 5 mol of N2O4 and allowed the system to reach equilibrium. If 2 mol of NO2 forms, the amount of N2O4 at equilibrium is (1)
A 1 mol
B 1.5 mol
C 3 mol
D 4 mol
b) Under different conditions, 25% of the moles of gas present at equilibrium is N2O4.If the total pressure of the system is 3 atm, the numerical value of the equilibrium constant Kp is (1)
A 9.00
B 6.75
C 3.00
D 0.15
in b part of the question how the answer is B? plz help
and this question is from edexcel chem unit 4 june 2015 ial question 6
Total pressure = sum of the partial pressures

Partial pressure = mole fraction x total pressure

pp(N2O4) = 0.25 x 3 = 0.75
pp(NO2) = 0.75 x 3 = 2.25

kp = (2.25)2/0.75 = 6.75
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mushriqite
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thankyou so muchhh!!!
(Original post by charco)
Total pressure = sum of the partial pressures

Partial pressure = mole fraction x total pressure

pp(N2O4) = 0.25 x 3 = 0.75
pp(NO2) = 0.75 x 3 = 2.25

kp = (2.25)2/0.75 = 6.75
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