Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:
$2 N O_{(g)} + B r_{2 (g)} ⇌ 2 NOB r_{(g)}$

When $0.087$ mole of $NO$ and $0.0437$ mole of $B r_{2}$ are mixed is a closed container at constant temperature, $0.0518$ mole of $NOB r$ is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide and bromine.

Question

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:

2 N O_{(g)} + B r_{2 (g)} ⇌ 2 NOB r_{(g)}

When

0.087

mole of

NO

and

0.0437

mole of

B r_{2}

are mixed is a closed container at constant temperature,

0.0518

mole of

NOB r

is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide and bromine.

Filo · Accepted Answer

2NO(g)​+Br2(g)​⇌2NOBr(g)​Initial moles of NO=0.087Initial moles of Br2​=0.0437Final moles of NO=2aFinal moles of Br2​=a∴ number of moles of NO at equilibrium =(0.087−2a)Number of moles of Br2​ at equilibrium =(0.0437−a)Given: 2a = 0.0518	herefore a = \dfrac{0.0518}{2}=0.0259Atequilibrium, Numberofmoles(amount)ofNO = 0.0870 - 0.0518 = 0.0352moleNumberofmoles(amount)ofBr_2 = 0.0437 - 0.0259 = 0.0178 mole

Question Text	Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below: $2 N O_{(g)} + B r_{2 (g)} ⇌ 2 NOB r_{(g)}$ When $0.087$ mole of $NO$ and $0.0437$ mole of $B r_{2}$ are mixed is a closed container at constant temperature, $0.0518$ mole of $NOB r$ is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide and bromine.
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