Question

For the reaction N₂ + 3H₂ → 2NH₃
if ∆[NH ]/∆t = 2 × 10⁻⁴ mol l⁻¹ s⁻¹, the value of ∆[H₂]/∆t
would be
(a) 1 × 10⁻⁴ mol L⁻¹ s⁻¹ (b) 3 × 10⁻⁴ mol L⁻¹ s⁻¹
(c) 4 × 10⁻⁴ mol L⁻¹ s⁻¹ (d) 6 × 10⁻⁴ mol L⁻¹ s⁻¹

Answer 1

Answer:

Here is your answer..

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Answer 2

answer : option (b) 3 × 10-⁴ mol l-¹ s-¹

for chemical reaction N₂ + 3H₂ → 2NH₃

-d[N₂]/dt = -1/3 d[H₂]/dt = 1/2 d[NH₃]/dt ......(1)

given, d[NH₃]/dt = 2 × 10-⁴ mol l-¹ s-¹

we have to find d[H₂]/dt

applying equation (1)

-1/3 d[H₂]/dt = 1/2 d[NH₃]/dt

⇒-1/3 d[H₂]/dt = 1/2 × 2 × 10-⁴ = 10-⁴

⇒d[H₂]/dt = -3 × 10-⁴ mol l-¹ s-¹

hence magnitude of d[H₂]/dt = 3 × 10-⁴ mol l-¹ s-¹

hence, option (b) is correct choice.

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