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

For the reaction `N_(2) + 3H_(2) to 2NH_(3)` if
`(Delta[NH_(3)])/(Deltat) = 2 xx 10^(-4) mol L^(-1)s^(-1)`, the value of `(-Delta[H_(2)])/(Deltat)` would be

A

`1 xx 10^(-4) mol L^(-1)s^(-1)`

B

`3 xx 10^(-4) mol L^(-1) s^(-1)`

C

`4 xx 10^(-4) mol L^(-1) s^(-1)`

D

`6xx 10^(-4) mol L^(-1) s^(-1)`

Text Solution

Verified by Experts

`1/2 (d[NH_3])/(dt) =- 1/3(d[H_2])/(dt)`
`:. -(d[H_2])/(dt)= 3/2xx(d[NH_3])/(dt)= 3/2xx 2xx 10^(-4)=3 xx 10^(-4)`
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