Question

The reaction n2(g)+o2(g)⇌2no(g) contributes to air pollution

Answer 1

Answer:

Let X amount has been decayed from reactant to product then

$N_{2} +O_{2}$↔2NO

T=0 0.8  0.20

T=$t_{equ}$   0.8-x 0.2-x 2x

$K_{c} =1.0*10^{-5}$

$K_{c} =\frac{[NO]^{2} }{[N_{2}][O_{2} ] }$

$1.0*10^{-5} =(2X)^{2} /[(0.8-X)(0.2-X)]$

If x is very small and then

0.8-x≈0.8

0.2-x≈0.2

$1.0*10^{-5} =(2x)^{2} /[(0.8)(0.2)]\\16*10^{-6}=4X^{2}  \\X^{2} =4*10^{-6}\\X=2*10^{-3}$

the amount of reactant and product at equilibrium is

$N_{2} =0.8-0.002=0.798\\O_{2} =0.2-0.002=0.198\\NO=2*2*10^{-3} =4*10^{-3}$