Nitric oxide reacts with Br2 and gives nitrosyl bromide as reaction given below: 2 NO (g)+ Br 2(g) leftharpoons 2 NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br 2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. The equilibrium amount of NO is

Question

Nitric oxide reacts with Br2 and gives nitrosyl bromide as reaction given below: 2 NO (g)+ Br 2(g) leftharpoons 2 NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br 2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. The equilibrium amount of NO is (A) 0.003 mol (B) 0.0178 mol (C) 0.0352 mol (D) 2.078 mol

Tardigrade Admin · Accepted Answer

2 NO (g)+ Br 2(g) leftharpoons 2 NOBr (g) beginarraylccl text Initial moles 0.087 0.0437 0 text Moles at eq. (0.087-2 x) (0.0437-x) 2 x endarray But 2x = 0.0518 .-. x = 0.0259 (nNo)eq = 0.087 - 0.0518 = 0.0352 mol

$0.003 m o$ l

$0.0178 m o l$

$0.0352 m o l$

$2.078 m o l$

$2 N O_{(g)} + B r_{2 (g)} ⇌ 2 NOB r_{(g)}$
$Initial moles Moles at eq. 0.087 (0.087 - 2 x) 0.0437 (0.0437 - x) 0 2 x$
But $2 x = 0.0518. - . x = 0.0259$
$(n_{N} o)_{e} q = 0.087 - 0.0518 = 0.0352 m o l$

0.003mol

0.0178mol

0.0352mol

2.078mol

2NO(g)​+Br2(g)​⇌2NOBr(g)​ Initial moles Moles at eq. ​0.087(0.087−2x)​0.0437(0.0437−x)​02x​ But 2x=0.0518.−.x=0.0259 (nN​o)e​q=0.087−0.0518=0.0352mol

$0.003 m o$ l

$0.0178 m o l$

$0.0352 m o l$

$2.078 m o l$

$2 N O_{(g)} + B r_{2 (g)} ⇌ 2 NOB r_{(g)}$
$Initial moles Moles at eq. 0.087 (0.087 - 2 x) 0.0437 (0.0437 - x) 0 2 x$
But $2 x = 0.0518. - . x = 0.0259$
$(n_{N} o)_{e} q = 0.087 - 0.0518 = 0.0352 m o l$