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 =

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.