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N(2)(g)+3H(2)(g)hArr2NH(3)(g) For the...

`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`
For the reaction intially the mole ratio was `1:3 ` of `N_(2):H_(2)`.At equilibrium 50% of each has reacted .If the equilibrium pressure is P, the parial pressure of `NH_(3)` at equilibrium is :

A

`P/3`

B

`P/4`

C

`P/6`

D

`P/8`

Text Solution

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The correct Answer is:
A
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