Question

An equilibrium mixture in a vessel of capacity 100 litre contain 1 mol N2, 2 mol O2 and 3 mol NO. Number of moles of O2 to be added so that at new equilibrium the conc. of NO is found to be 0.04 mol/lit.:

• A. (101/18)
• B. (101/9)
• C. (202/9)
• D. None of these.

Answer 1

Answer: (A)

(101/18)

Answer 2

N2 (g) + O2 (g) 2NO (g)

1mole 2mole 3mole

KC =$\frac { ( 3 ) ^ { 2 } } { 1 \times 2 }$ = $\left( \frac { 9 } { 2 } \right)$.

Let a mole of O2 is added, Then,

N2 (g) + O2 (g) 2NO (g)

1mole 2mole 3mole

t=0 1 (2 + a) 3

(1–x) (2 + a)–x (3 + 2x)

[NO] = $\left\lbrack \frac{3 + 2x}{100} \right\rbrack$ = 0.04 ;

(3 + 2x) = 4.

2x = 1, x = 0.5.

KC = $\frac{(3 + 2x)^{2}}{(1 - x)(2 + a - x)} = \frac{9}{2}$

$K_{C} = \frac{(4)^{2}}{0.5\left\lbrack (1.5) - a \right\rbrack} = \frac{9}{2}$.

$= \frac{16}{0.5(1.5 + a)} = \frac{9}{2}$.

$= \frac{35}{4.5} = \lbrack 1.5 + a\rbrack$

$7.11 = 1.5 + a$

$a = \frac{101}{8} = 5.61$