Question

The equilibrium constant (Kc) for the 3
reaction N2(g) +O28) 2NO(g) at
temperature T is 4x10-4. The value of K.
for the reaction, NO(g) = N216) +2026)
at the same temperature is
[2012]
(A) 2.5x102
(C) 50.0
(B) 4x10-4
(D) 0.02
for the renctions​

Answer 1

Explanation:

The value of K_cK

c

for the reaction NO(g)\rightarrow \frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)NO(g)→

2

1

N

2

(g)+

2

1

O

2

(g) at the same temperature is 5\times 10^{-3}5×10

−3

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}K

eq

For a general chemical reaction:

N_2(g)+O_2(g)\rightarrow 2NO(g)N

2

(g)+O

2

(g)→2NO(g)

The expression for K_{eq}K

eq

is written as:

K_{c}=\frac{[NO]^2}{[N_2]^1[O_2]^1}K

c

=

[N

2

]

1

[O

2

]

1

[NO]

2

K_{c}=4\times 10^4K

c

=4×10

4

For the reaction :

NO(g)\rightarrow \frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)NO(g)→

2

1

N

2

(g)+

2

1

O

2

(g)

That is the above reaction is reversed and halved, the new equilibrium constant is:

K'_{c}=\frac{1}{\sqrt {K_c}}=\frac{1}{\sqrt {4\times 10^4}}=\frac{1}{200}=5\times 10^{-3}K

c

′

=

K

c

1

=

4×10

4

1

=

200

1

=5×10

−3