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In a catalytic conversion of N(2) to NH(...

In a catalytic conversion of `N_(2)` to `NH_(3)` by Haber's process, the rate of reaction was expressed as change in the concentration of ammonia per time is `40 xx 10^(-3) mol L^(-1) s^(-1)`. If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is (in mol `L^(-1) s^(-1)`)

A

`60xx10^(-3)`

B

`20xx10^(-3)`

C

`1.200`

D

`10.3xx10^(-3)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(-d(N_(2)))/(dt)=-(1)/(3)(d(H_(2)))/(dt)=(1)/(2)(d(NH_(3)))/(dt)=(3)/(2)xx40xx10^(-3) = 60xx10^(-3)`
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