Question

Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature, are given below. reaction (1): N2(g) + O2(g) 2 NO(g); Kc = 1.54e-31 reaction (2): N2(g) + 1/2 O2(g) N2O(g); Kc = 2.61e-24 Using this set of data, determine the equilibrium constant for the following reaction, at the same temperature. reaction (3): N2O(g) + 1/2 O2(g) 2 NO(g)

Answer 1

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   $N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)$  ; Kc = 1.54e - 31

2)   $N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)$  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     $N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       $N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)$

      $N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$

   we get ,

                  $N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)$  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24