Question

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

Answer 1

The value of – d [H₂ ]/dt

would be (d) 3 × 10⁻⁴mol L⁻¹ s⁻¹

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

or 1/2 d[NH₃]/dt = -1/3 d[H₂]/dt

or 1/2 * 2 × 10⁻⁴ = - 1/3 [H₂ ]/dt

or 3 × 10⁻⁴mol L⁻¹ s⁻¹ = – d [H₂ ]/dt

2. Thus (d) is the correct answer

Answer 2

answer : option (d) 3 × 10⁴ mol L-¹ s-¹

for the reaction, N₂ + 3H₂ → 2NH₃

rate of disappearance of reactant = rate of appearance of product = 1/stoichiometric coefficient × d[reactant/product]/dt

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

here given, d[NH₃]/dt = 2 × 10⁻⁴ mol L⁻¹ s⁻¹

and we have to find d[H₂]/dt

so, -1/3 d[H₂]/dt = 1/2 d[NH₃]/dt [from eq (1) ]

⇒-1/3 d[H₂]/ft = 1/2 × 2 × 10⁴ = 10⁴

⇒-d[H₂]/dt = 3 × 10⁴ mol L-¹ s-¹

hence option (d) is correct choice.

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