Question

Explain the Relation between K_C and K_P.

Answer 1

Consider a general reversible reaction:

aA_(g) + bB_(g) ⇌ cC_(g) + dD_(g)

The equilibrium constant (K_P) 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,

P_AV = n_ART

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

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

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

Similarly, for other components, P_B = [B]RT, P_C = [C]RT, P_D = [D]RT

Now substituting equations for P_A, P_B, P_C, P_D 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