Question

don't spam answer correctly​

Answer 1

"Answer:

1. -0.02 is the correct answer.

Explanation:

$-\frac{1}{3} \frac{d[H_2]}{dt} = 0.01M/min = -1.667 X 10^{-4}$ M/sec

$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{db}$

$-1.667 X 10^{-4}$ =$\frac{1}{2}\frac{d[NH_3]}{dt}$ = $2 * (-1.667 X 10^{-4} M/sec) = -3.34 X 10^{-4} M/s$

Therefore we can say the Rate of NH3 formed is =$-3.34 X 10^{-4} M/s$

Rate of formation is M/min = $-3.34 X 10^{-4} M/s * 60 sec = -0.02 M/min$."

Answer 2

Answer

i don't know the concept but u can prefer this sor.ry i couldn't help

$\frac{-d[N2]}{ddt} = \frac{1}{2} \frac{d[NH3]}{dt}\\\\ \frac{d[NH3]}{dt}$

DIDI MA IN$TA USE NAHI KAR TA AP MUJA YA F.B PA INB0X KARO GA ?