Question

Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H20).

Answer 1

Answer: The equilibrium concentration of $CH_4$ , expressed in scientific notation is $5.9\times 10^{-2}M$

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

                          $CO(g)+3H_2(g)\rightleftharpoons CH_4(g)+H_2O(g)$

At eqm. conc.     (0.30) M     (0.10) M      (x) M   (0.020) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CH_4]\times [H_2O]}{[CO]\times [H_2]^3}$

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

$3.90=\frac{x\times (0.020)}{(0.30)\times (0.10)^3}$

By solving the term 'x', we get :

x = 0.059 M=  $5.9\times 10^{-2}M$

Thus, the concentrations of $CH_4$ at equilibrium is :

Concentration of $CH_4$ = (x) M  = $5.9\times 10^{-2}M$

The equilibrium concentration of $CH_4$ , expressed in scientific notation is $5.9\times 10^{-2}M$