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Question
Derive the relation between Kp and Kc.
Solution
Let us consider the general reaction in which all reactants and products are ideal gases.
xA + yB ⇌ lC + mD
The equilibrium constant, Kc is,
`"K"_"c" = (["Cl"]^"l"["D"]^"m")/(["A"]^"x"["B"]^"y")` ...(1)
and Kp is,
`"K"_"p" = ("p"_"C"^"l" xx "p"_"D"^"m")/("p"_"A"^"x" xx "p"_"B"^"y")` ...(2)
The ideal gas equation is,
PV = nRT
P = `"n"/"V"`RT
since, Active mass = molar concentration = n/V
P = active mass × RT
Based on the above expression the partial pressure of the reactants and products can be expresssed as,
PXA = [A]x[RT]x
PYB = [A]y[RT]y
PlC = [A]l[RT]l
PmD = [A]m[RT]m
On substituting in Eqn.(2).,
`"K"_"p" = (["C"]^"l"["RT"]^"l" ["D"]^"m" ["RT"]^"m")/(["A"]^"x" ["RT"]^"X" ["B"]^"y" ["RT"]^"y")` .....(3)
`"K"_"p" = (["C"]^"l"["RT"]^"l" ["D"]^"m" ["RT"]^("l + m"))/(["A"]^"x"["B"]^"y" ["RT"]^("x + y"))`
`"K"_"p" = (["C"]^"l"["D"]^"m")/(["A"]^"x"["B"]^"y") ["RT"]^(("l + m") - (x + y))` ....(4)
By comparing equation (1) and (4),
we get
Kp∆ng =Kc (RT) …………..(5)
Where,
∆ng is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
The following relations become immediately obvious.
when ∆ng = 0.
Kp = Kc (RT)0 = Kc
Example:
\[\ce{H2(g) + l_2(g) <=> 2HI(g)}\]
\[\ce{N2 + O2 <=> 2NO(g)}\]
when ∆ng = +ve
Kp = Kc(RT)+ve
Kp > Kc
Example:
\[\ce{2NH3(g) <=> N2(g) + 3H2(g)}\]
\[\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}\]
When ∆ng = -ve
Kp = Kc(RT)-ve
Kp < Kc
Example:
\[\ce{2H2(g) + O2(g) <=> 2H2O(g)}\]
\[\ce{2SO2(g) + O2(g) <=> 2SO3(g)}\]
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