In an indoor atmosphere, for No_2 the value of the first-order rate constant has

Question

In an indoor atmosphere, for No_2 the value of the first-order rate constant has been estimated to be 1.28 h^-1. calculate its residence time. The equilibrium constant for the reaction has a value K_c = 4.65 times 10^-3 L mol^-1 at 25 degree C and the corresponding standard enthalpy change is delta H degree = +57 kJ. Calculate the equilibrium concentration of N_2o_4 in an atmosphere where the concentration of NO_2 is 200 mug m^-3. If the temperature were to increase would you except the relative concentration of N_2o_4 to increase or decrease?

Explanation / Answer

18)

consider the given equilibrium

N204 ----> 2 N02

the equilibrium constant is given by

Kc = [N02]^2 / [N204]

given

[N02] = 200 ug / m3

we know that

1 m3 = 1000 L

so

[N02] = 200 ug / 1000 L

[N02] = 0.2 ug / L

now

1 ug = 10-6 g

so

[N02] = 0.2 x 10-6 g / L

[N02] = 2 x 10-7 g / L

now

we know that

concentration in mol / L = conc in g/ L / molar mass

molar mass of N02 = 46 g / mol

so

[N02] = 2 x 10-7 / 46 mol / L

[N02] = 4.347826 x 10-9 mol / L

now

Kc = [N02]^2 / [N204]

4.65 x 10-3 = [4.347826 x 10-9]^2 / [N204]

[N204] = 4.0653 x 10-15 mol / L

now

1 mole of N204 = 92 g

so

[N204] = 4.0653 x 10-15 x 92 g / L

[N204] = 3.74 x 10-13 g / L

[N204] = 3.74 x 10-7 x 10-6 g / L

[N204] = 3.74 x 10-7 ug / L

[N204] = 3.74 x 10-7 ug / 10-3 m3

[N204] = 3.74 x 10-4 ug / m3

so

the equilibrium concentration of N204 is 3.74 x 10-4 ug / m3

now

given that

dH = +ve

so

N204 + heat ---> 2 N02

so

according to Le Chatlier principle

upon increase in temperature

the relative concentration of N204 will decrease

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.