FOR PRACTICE 15.6 The reaction
of CH4 in Example 15.6 is carried
out at a different temperature
with an initial concentration of
[CH] = 0.087 M. At equilibrium,
the concentration of His 0.012 M.
Find the equilibrium constant at
this temperature.

The equilibrium constant will be "1.1×10⁻⁶ m²".

Explanation:

CH₄ has a varying temperatures response with such an initial concentration of [CH₄] = 0.087 m.

At equilibrium, concentration of H₂ will be 0.012 m.

2CH₄           ⇄         C₂H₂ + 3H₂

(0.087 m)                  (0)       (0)

(0.087-2x)                 (x)        (3x)

[H₂] = 0.012 m

⇒  3x = 0.012

⇒    x = 0.004m

Now,

On putting the value of "x" in the above expression, we get

⇒ (0.087-0.004×2)          (0.004)           (0.012 m)

⇒ (0.079 m)                     (0.004)            (0.012 m)

As we know,

On putting the values in the above formula, we get

⇒  

⇒        

⇒        

what does the table look like

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.