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In the synthesis of ammonia by Harber's ...

In the synthesis of ammonia by Harber's process. If 60 moles of ammonia is obtained in one hour, then the rate of disappearence of nitrogen is:

A

`30` mol/min

B

`6` mol/ min

C

`0.5` mol/min

D

`60` mol/min

Text Solution

Verified by Experts

The correct Answer is:
C
`N_(2)+3H_(2)=2NH_(3)`
`"rate"=-(d[N_(2)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)=(1)/(2)(d[NH_(3)])/(dt)`
Rate of disappearance of `N_(2)`
= `(1)/(2)` of rate of formation of `NH_(3)`
= `(1)/(2)xx(60"mole")/(1"hour")=(1)/(2)xx(60)/(60)` mole/ minute
`0.5` mole/minute .
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