$(a)\;1\;atm\qquad(b)\;2\;atm\qquad(c)\;3\;atm\qquad(d)\;4\;atm$

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$2NH_3 \rightarrow N_2 +3H_2$

at same P and T

Number of moles after reaction $n_2 = 2\times$ number of moles before reaction $(n_1)$

$\therefore P\propto n$

$\therefore \large\frac{P_1}{P_2} = \large\frac{n_1}{n_2}$

$\;\;\;\;\;\;\;\;\;\;=\large\frac{1}{2}$

Pressure after reaction = 2 atm

Hence answer is (b)

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