Question

for N2+3H2=2NH3 rate of disappearance of H2 is 0.01M/min the amount of NH3 formed at yhat instant
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Answer 1

Explanation:

[tex]begin{array}{l} N_{2}+3 H_{2} \rightarrow 2 N H_{3} \\ -\frac{d\left[H_{2}\right]}{d t}=0.01 M/min=0.3 \times 10^{-4} M s^{-1} \\  \text {Rate }=\frac{1}{3} \frac{-d\left[H_{2}\right]}{d t}=+\frac{1}{2} \frac{d\left[N H_{3}\right]}{d t} \\ =\frac{d\left[N H_{3}\right]}{d t}=\frac{2}{3} \frac{-d\left[H_{2}\right]}{d t} \\ =\frac{d\left[N H_{3}\right]}{d t}=\frac{-2}{3} \times\left(0.3 \times 10^{-4}\right) \\  \frac{d\left[N H_{3}\right]}{d t}=-0.2 \times 10^{-4} M s^{-1} \end{array}[/tex]

Answer 2

Answer:

0.1132 gm

Explanation:

1. d[nh3]/dt=-2/3dh2/dt.