JEE Main 2019 — Chemical Equilibrium Question with Solution
From: JEE Main 2019 (Online) 11th January Morning Slot
Question
Consider the reaction
N2(g) + 3H2(g) 2NH3(g)
The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium)
N2(g) + 3H2(g) 2NH3(g)
The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium)
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
N2(g) + 3H2(g) 2NH3(g) ; Keq = Kp
Write this equation reverse way,
2NH3(g) N2(g) + 3H2(g) ; Keq =
At equillibrium
PTotal = PNH3 + PN2 + PH2
= PNH3 + p + 3p
(As PNH3 << Ptotal so we can ignore PNH3)
PTotal = 4p
p =
Formula of
Keq = =
=
= Kp 27
PNH3 =
=
Write this equation reverse way,
2NH3(g) N2(g) + 3H2(g) ; Keq =
| 2NH3(g) | ⇌ | N2(g) | + | 3H2(g) | |
|---|---|---|---|---|---|
| At t = 0 | Po | 0 | 0 | ||
| At t = teq | PNH3 | p | 3p |
At equillibrium
PTotal = PNH3 + PN2 + PH2
= PNH3 + p + 3p
(As PNH3 << Ptotal so we can ignore PNH3)
PTotal = 4p
p =
Formula of
Keq = =
=
= Kp 27
PNH3 =
=
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