Discussion Forum



1)

For the reaction N2 + 3H2 → 2NH3 , If $\frac{Δ[NH_{3}]}{Δt} = 2\times 10^{-4} mol/ls$  then value of $\frac{-Δ\left[H_{2}\right]}{Δt}$ would be


A) $1\times 10^{-4}mol L^{-1}S^{-1}$

B) $3\times 10^{-4}mol L^{-1}S^{-1}$

C) $4\times 10^{-4}mol L^{-1}S^{-1}$

D) $6\times 10^{-4}mol L^{-1}S^{-1}$

Answer:

Option B

Explanation:

 N2 + 3H2 → 2NH3

$\frac{-Δ[N_{2}]}{Δt}$ = $\frac{-1}{3}\frac{Δ\left[H_{2}\right]}{Δt}$

= $\frac{1}{2}\frac{Δ[NH_{3}]}{Δt}$

$\frac{-Δ\left[H_{2}\right]}{Δt}$ = $\frac{3}{2}$× $\frac{Δ[NH_{3}]}{Δt}$

=$\frac{3}{2}$× $2\times 10^{-4} mol/ls$

= $3\times 10^{-4}mol L^{-1}S^{-1}$