Question

The formation of methane gas occurs at 1200 k in following reaction co(g)

3h2(g) <----> ch 4(g) + h2o (g)
1 mole of co, 3 moles of h2 are placed in a 10 l reaction vessel. if the reaction is found to contain 0.387 mole of h2o, then the h2 amount (in moles) at equilibrium is:____. question 5 options:
a. 0.387
b. 0.774
c. 0.912
d. 1.839
e. 0.613

Answer 1

The amount of H₂ at equilibrium, given that 0.387 moles of H₂O were formed in the reaction CO(g) + 3H₂(g) ↔ CH₄(g) + H₂O(g), is 1.839 moles. This is determined by subtracting the moles of H₂ consumed (1.161 moles) from the initial moles of H₂ (3 moles).

Step-by-step explanation:

The student has asked for the amount of H₂ (hydrogen gas) at equilibrium given the formation of methane gas through the reaction CO(g) + 3H₂(g) ↔ CH₄(g) + H₂O(g).

The reaction starts with 1 mole of CO and 3 moles of H₂ in a 10 L reaction vessel at 1200 K.

If there are 0.387 moles of H₂O at equilibrium, we can determine the moles of H₂ left at equilibrium.

Initial moles of H₂: 3 moles
Since 0.387 moles of H₂O are formed, we can infer that the same amount of CH₄ was also formed (as the molar ratio between CH₄ and H₂O is 1:1). This means 0.387 moles of CO and 3 × 0.387 moles of H₂ are consumed.

H₂ consumed = 0.387 moles × 3 = 1.161 moles
H₂ remaining = initial moles of H₂ - H₂ consumed
H₂ remaining = 3 moles - 1.161 moles = 1.839 moles

Thus, option D, 1.839 moles, is the correct amount of H₂ at equilibrium.