200 cm3 of hydrogen iodide gas (hi) is allowed to dissociate and come to equilibrium. the mixture at equilibrium contained 350 cm3 of gaseous iodine. calculate k, for the reaction 2hi(g)h(g)+ ig) (assume that 1 mol of a gas has a volume of 35 000 cm under the conditions in this experiment.)

Answer : The value of 'K' equilibrium constant is, 0.489

Solution :  Given,

Volume of =

Volume of at equilibrium =

Volume of gas for 1 mole of gas =

First we have to calculate the moles of and .

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

Initially moles     0.0057         0       0

At equilibrium  (0.0057-2x)     x        x

The expression of will be,

Let the total volume be 'V'.

As per question,

Moles of at equilibrium = x = 0.01

Now put the values of 'x' in the above expression, we get:

Therefore, the value of 'K' equilibrium constant is, 0.489

explanation:

newton's third law, for every action there is an equal and opposite reaction.