The reaction 2CH4 (g) ⇌C2H2 (g) + 3H2 (g) has an equilibrium constant of K = 0.154. If 6.70 mol of CH4, 4.05 mol of C2H2, and 10.05 mol of H2 are added to a reaction vessel with a volume of 5.20 L, what net reaction will occur?

Q>> K

the reaction will proceed to the left

Explanation:

Step 1: Data given

K = 0.154

Number moles CH4 = 6.70 moles

Number of moles C2H2 = 4.05 moles

Number of moles H2 = 10.05 moles

Volume = 5.20 L

Step 2: Calculate concentration of CH4

Concentration = moles / volume

[CH4] = 6.70 mol / 5.20 L = 1.29 M

Step 3: Calculate concentration C2H2

[C2H2] = 4.05 mol / 5.20 L = 0.779 M

Step 3: Calculate concentration H2

[H2] = 10.05 mol / 5.20 L = 1.93 M

Q = [C2H2] [H2]³ / [CH4]²

Q = 0.779 (1.93) ³ / (1.29) ² = 3.37 >> K

Q>> K

the reaction will proceed to the left