Question

Question 7.9 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:

2NO(g) + Br2(g) <---> 2NOBr(g)

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.

Class XI Equilibrium Page 225

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2NO(g) + Br2(g) < = > 2NOBr(g) \\ at \: t = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: 0.087 \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0.0437 \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ at \: eqlm \: \: \: \: \: \: (0.087 - 2x) \: \: \: \: \: \: \: \: \: \: \: \: (0.0437 - x) \: \: \: \: \: \: \: \: \: \: \: \: 2x$
Given, moles of NOBr at equilibrium = 0.518
e.g., 2x = 0.0518
x = 0.0259
moles of NO at equilibrium = 0.087 - 2x
= 0.087 - 2 × 0.0259
= 0.087 - 0.0518
= 0.0352 mol

moles of Br2 at equilibrium = 0.0437 - x
= 0.0437 - 0.0259
= 0.0178 mol

Answer 2

THANKS FOR ASKING

THE REQUIRED ANSWER IS IN ABOVE ATTACHED PICTURE.