Question

a) Give the mathematical equilibrium constant expression Kc and Kp (not the value!) for the reaction

2N2O5(g) ↔ 4NO2 (g) + O2(g)

b) At 250oC the concentrations of N2O5, NO2 and O2 are 0.100 mol/L, 0.0283 mol/L and 0.0105 mol/L. Calculate the values for Kc and Kp

Answer 1

(a) 2N₂O₅(g) ↔ 4NO₂ (g) + O₂(g)

equilibrium constant K_c = ( [ O₂(g)] [NO₂ (g)]⁴ ) / [N₂O₅(g)]²

equilibrium constant K_p = ( p_O2(g) x p⁴_NO2(g) ) / p²_N2O5(g)

The mathemetical relation between K_p & K_c is K_p = K_c x (RT) ⁿ

(b)Given the Equilibrium concentrations are as follows:

[ O₂(g)] = 0.0105 mol / L

[NO₂ (g)] = 0.0283 mol / L

[N₂O₅(g)] = 0.100 mol / L

equilibrium constant K_c = ( [ O₂(g)] [NO₂ (g)]⁴ ) / [N₂O₅(g)]²

                                      = ( 0.0105 x (0.0283)⁴ ) / 0.100²

                                      = 6.73x10^-7

We know that K_p = K_c x (RT) ⁿ

Where

R = gas constant = 0.0821 L-atm/ mol-K

T = Temperature = 250 ^oC = 250+273 = 523 K

n = change in number of moles

       = number of moles of gaseous products - number of moles of gaseous reactants

      = (4+1) - 2

      = 3

Plug the values we get

K_p = K_c x (RT) ⁿ

K_p = 6.73x10^-7 x (0.0821x523) ³

      ⁼ 0.053