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For the reaction N(2)+3H(2) rarr 2NH(3),...

For the reaction `N_(2)+3H_(2) rarr 2NH_(3)`, if `(d[NH_(3)])/(dt).=4xx10^(-4)` mol `L^(-1)s^(-1)`, the value of `(-d[H_(2)])/(dt)` would be

A

0.02

B

50

C

0.06

D

0.04

Text Solution

Verified by Experts

The correct Answer is:
C
`(-d[N_2])/(dt) = - 1/3 (d[H_2])/dt`
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