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

C₂H₂ + 3H₂ ⟶ 2CH₄

Step-by-step 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 = \frac{\text{[C$_(2)$H$_(2)$][H$_(2)$]}^(3)}{\text{[CH$_(4)$]}^(2)} = ( 0.700* 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