Question

When 0.112 mol of no and 18.22 g of bromine are placed in 1.00 l reaction vessel and sealed, the mixture is heated to 350 k and equilibrium is established. 2no(g) + br2(g) 2nobr(g) if the equilibrium [nobr] concentration is 0.0824m,

Answer 1

The question is incomplete. complete question is ;

When 0.112 mol of no and 18.22 g of bromine are placed in 1.00 l reaction vessel and sealed, the mixture is heated to 350 k and equilibrium is established.

$2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)$

if the equilibrium [NOBr] concentration is 0.0824 M, then find equilibrium constant.

The value of the equilibrium constant is 106.45.

Explanation:

Concentration before attaining equilibrium :

Moles of NO =0.112 mol

$[NO]=\frac{0.112 mol}{1 L}=0.112 M$

Moles of bromine = $\frac{18.22 g}{160 g/mol}=0.1139 mol$

$[Br_2]=\frac{0.1139 mol}{1 L}=0.1139 M$

$[NOBr]=0$

$2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)$

Initially

0.112 M           0.1139 M            0

At equilibrium

(0.112- 2 × 0.0824 )M    (0.1139 -(0.0824/2) )M      0.0824 M

0.0296 M                 0.0728 M       0.0824 M

Expression of equilibrium constant is given by ;

$K_c=\frac{[NOBr]^2}{[NO]^2[Br_2]}$

$K_c=\frac{(0.0824 M)^2}{(0.0296 M)^2\times (0.0728 M)}= 106.45$

The value of the equilibrium constant is 106.45.

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