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

For the reaction, `N_(2)+3H_(2) rarr 2NH_(3)` the rate `(d[NH_(3)])/(dt)=2xx10^(-4) M s^(-1)`. Therefore the rate `-(d[N_(2)])/(dt)` is given as:

A

`10^(-4) M sec^(-1)`

B

`10^(4) M sec^(-1)`

C

`10^(-2) M sec^(-1)`

D

`10^(-4) sec^(-1) M^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
a
`(-d[N_(2)])/(dt)=1/2(d[NH_(3)])/(dt)`
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