Consider the reaction CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g) Kc = 7.7×10–23 at 25 °C. 1.50 mol CH4 and 2.50 mol H2O are added to an empty 2.00-L container at 25 °C and allowed to reach equilibrium. What are the equilibrium amounts (in moles) of each species? 1.5

Answer : The moles of , , and at equilibrium is, 0 mol, 1 mol, 1.5 mol and 4.5 mol respectively.

Explanation :

First we have to calculate the concentration of

and,

The given chemical reaction is:

Initial conc.     0            0             0.75        1.25

At eqm.            x            3x         (0.75-x)     (1.25-x)

The expression for equilibrium constant is:

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

x = -2.19317 × 10¹⁰

x = 0.75

x = 1.25

x = 2.19317 × 10¹⁰

We are accepting value of x = 0.75 while all the values of x are neglecting because equilibrium concentration can not be more than initial concentration.

Equilibrium concentration of = (0.75-x) = (0.75-0.75) = 0 M

Equilibrium concentration of = (1.25-x) = (1.25-0.75) = 0.50 M

Equilibrium concentration of = x = 0.75 M

Equilibrium concentration of = 3x = 3(0.75) = 2.25 M

Now we have to calculate the moles of each species in terms of mole.

Moles of at equilibrium =

Moles of at equilibrium =

Moles of at equilibrium =

Moles of at equilibrium =

answer; /// the amount of material dissolved in a measure of liquid(concentration) the above statement is (true); /// good luck