1 mole of N2 and 3 moles of H2 produces NH3 at equilibrium calculate the concentration of NH3 at equilibrium if Kc = 4/27
Answers (2)
N2+3H2<->2NH3
Then
N2=1-X
H2=3-3X
NH3=2X
THEN K=4/27
AND
X/1-X^2=1
Then
N2=1-X
H2=3-3X
NH3=2X
THEN K=4/27
AND
X/1-X^2=1
Comments (0)
22nd Nov, 2018
The Reaction can be represented as
N2+3H2=2NH3
Let x mol of N2 Got reacted till equilibrium is reached, then According to stoichiometry, Moles of the component Present at equilibrium.
N2=1-x
H2=3-3x
NH3=2x
Then Equilibrium Constant
K=(2x)^2/27(1-x)^3=4/27
therefore x/(1-x)^2=1
Now You can calculate the X and hence the corresponding concentration at Equilibrium.
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