Question

The elementary reaction 2H2O(g) (forward, reverse arrows) 2H2(g) + O2 proceeds at a certain temperature until the partial pressure of H2O, H2 and O2 reach 0.0450atm, 0.00300atm and 0.00650atm respectively, What is the value of the equilibrium constant at this temperature. I need help please!

Answer 1

Let; We have given a reaction as;

2H₂O_(g) 2H_2(g) + O_2(g)

Also, We have given the partial pressures as equilibrium as;

P_H2O = 0.0450 atm

P_H2 =0.00300 atm

P_O2 = 0.00650 atm

Now; From the given reaction; the expression for equilibrium constant is given as;

K_p = [(P_H2)² (P_O2)] / (P_H2O)²

Now, Putting the values of equilibrium pressure in the above expression, We get;

K_p = [ (0.00300)² x (0.00650)] / [0.0450]²

  = (5.85x10^-8)/(2.025x10^-3)

K_p = 2.88 x10^-5

The value of the equillibrium constant is 2.88 x10^-5.