# Explain the Relation between KC and KP. - Chemistry

Explain the Relation between KC and KP.

#### Solution

Consider a general reversible reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

The equilibrium constant (KP) in terms of partial pressure is given by equation:

"K"_"P" = (("P"_"C")^"c"("P"_"D")^"d")/(("P"_"A")^"a"("P"_"B")^"b")   .....(1)

For a mixture of ideal gases, the partial pressure of each component is directly proportional to its concentration at constant temperature.

For component A,

PAV = nART

"P"_"A" = "n"_"A"/"V" xx "RT"

"n"_"A"/"V" is molar concentration of A in mol dm-3

∴ PA = [A]RT   where, [A] = "n"_"A"/"V"

Similarly, for other components, PB = [B]RT, PC = [C]RT, PD = [D]RT

Now substituting equations for PA, PB, PC, PD in equation (1), we get

"K"_"P" = (["C"]^"c"("RT")^"c"["D"]^"d"("RT")^"d")/(["A"]^"a"("RT")^"a"["B"]^"b"("RT")^"b")

∴ "K"_"P" = (["C"]^"c" ["D"]^"d"("RT")^"c+d")/(["A"]^"a"["B"]^"b"("RT")^"a + b")

∴ "K"_"P" =(["C"]^"c" ["D"]^"d")/(["A"]^"a"["B"]^"b") xx ("RT")^(("c + d")-("a + b"))

∴ "K"_"P" =(["C"]^"c" ["D"]^"d")/(["A"]^"a"["B"]^"b") xx ("RT")^(Delta"n")

∴ But, "K"_"C" =(["C"]^"c" ["D"]^"d")/(["A"]^"a"["B"]^"b")

"K"_"P" = "K"_"C"("RT")^(Delta"n")

where Δn = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation.

R = 0.08206 L atm K–1 mol–1

Concept: Law of Mass Action and Equilibrium Constant
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#### APPEARS IN

Balbharati Chemistry 11th Standard Maharashtra State Board
Chapter 12 Chemical Equilibrium
Exercises | Q 3. (B) | Page 189
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