$N_2\left(g\right)$ and $H_2\left(g\right)$ are taken in a vessel in mass ratio of  $7:1$. The only reaction  $N_2+2H_2\rightleftharpoons N_2{H}_4\left(g\right)$ occurs. Pressure due to  $N_2{H}_4$ at equilibrium is $0.2$ times of total pressure. Then at equilibrium, which of the following is not true.

Equilibrium of N₂ + 2H₂ ⇌ N₂H₄

Statement: “All the reactants are completely converted — no N₂ or H₂ remain at equilibrium.”

Take 1 mol N₂ and 2 mol H₂ (mass ratio 7:1 → mole ratio 1:2).
Let extent of reaction = ξ.
At equilibrium:
N₂ = 1 − ξ
H₂ = 2 − 2ξ
N₂H₄ = ξ
Total moles = 3 − 2ξ

Given partial pressure of N₂H₄ = 0.2 of total pressure → mole fraction:

ξ / (3 − 2ξ) = 0.2
Solve: ξ = 3/7 ≈ 0.4286

So at equilibrium:

N₂ = 1 − 3/7 = 4/7 (not zero)
H₂ = 2 − 6/7 = 8/7 (not zero)
N₂H₄ = 3/7

Conclusion: N₂ and H₂ are not completely consumed. The statement is false.