Question

1 mole of NO2 and 2 moles ofCO are enclosed in 1 litre vessels to attain the following equlibrium NO2+CO=NO+CO2 it was estimated that the equilibrium 25%of initial amount of CO is consumed the equlibrium constant

Answer 1

Answer:

Please refer to the picture attached!

(Note: 25% of CO (2 mol.) = 0.5 mol)

Answer 2

Answer: The equilibrium constant is 0.33

Explanation:

Initial moles of  $NO_2$ = 1 mole

Volume of container = 1 L

Initial concentration of $NO_2=\frac{moles}{volume}=\frac{1mole}{1L}=1M$  

Initial moles of  $CO$ = 2 mole

Volume of container = 1 L

Initial concentration of $CO=\frac{moles}{volume}=\frac{2mole}{1L}=2M$  

equilibrium concentration of $CO=2M-\frac{25}{100}\times 2=1.5$

The given balanced equilibrium reaction is,

         $NO_2(g)+CO(g)\rightleftharpoons NO(g)+CO_2(g)$

Initial conc.        1M     2M                    0          0

At eqm. conc.    (1-x) M   (2-x) M         (x) M       (x)M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO][CO_2]}{[NO_2][CO]}$

$K_c=\frac{x\times x}{(2-x)(1-x)}$

We are given :

(2- x) = 1.5

x= 0.5

Now put all the given values in this expression, we get :

$K_c=\frac{(0.5)(0.5)}{(1-0.5)(2-0.5)}$

$K_c=0.33$

Thus the value of the equilibrium constant is 0.33