`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)`
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The correct Answer is:
a
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