Question

The reaction 2CH4(g)⇌C2H2(g)+3H2(g) has an equilibrium constant of K = 0.154. If 6.30 mol of CH4, 4.20 mol of C2H2, and 11.15 mol of H2 are added to a reaction vessel with a volume of 6.00 L , what net reaction will occur?

Answer 1

Answer:

C₂H₂ + 3H₂ ⟶ 2CH₄  

Explanation:

The initial concentrations are:

[CH₄] = 6.30 ÷ 6.00 =   1.05  mol·L⁻¹

[C₂H₂] = 4.20 ÷ 6.00 = 0.700 mol·L⁻¹

   [H₂] = 11.15 ÷  6.00 =  1.858 mol·L⁻¹

                2CH₄ ⇌ C₂H₂ + 3H₂

I/mol·L⁻¹:    1.05     0.700   1.858

$Q = \dfrac{\text{[C _{2} H _{2} ][H _{2} ]}^{3}}{\text{[CH _{4} ]}^{2}} = \dfrac{ 0.700\times 1.858^{3}}{1.05^{2}}= 4.07$

Q > K

That means we have too many products.

The reaction will go to the left to get rid of the excess products.

C₂H₂ + 3H₂ ⟶ 2CH₄