When N_2O_5(g) is heated, it dissociates into N_2O_3(g) and O_2(g) according to

Question

When N_2O_5(g) is heated, it dissociates into N_2O_3(g) and O_2(g) according to the reaction: N_2O_6(g) N_2O_3(g) + O_2(g) K_c = 7.75 at a given temperature The N_2O_3(g) dissociates to give N_2O(g) and O_2(g) according to the reaction: N_2O_6(g) N_2O(g) + O_2(g) K_c = 4.00 at the same temperature When 4.00 mol of N_2O_2(g) is heated in a 1.00 L reaction vessel to this temperature. the concentration of O_2(g) at equilibrium is 4.50 mol/L. Find the concentration of all the other species in the system at equilibrium.

Explanation / Answer

..........N2O5(g) <.......> N2O3(g) + O2(g)

...at eq...(4.0-x).................x.............x

Kc = [N2O3][O2]/[N2O5]

7.75 = x*x/(4.0-x)

31-7.75x =x^2

x^2 + 7.75x - 31 = 0

by solving the above quadratic equation we can get the x value is 2.91

so [N2O3] = 2.91 and [O2] = 2.91

..............N2O3<.......> N2O + O2

at eq.......(2.91-x)............x........(2.91+x)

Kc = [N2O][O2]/[N2O3]

4 = x*(2.91)/(2.91-x)

11.64 -4x = 2.91x

6.91x = 11.64

X = 1.68

at equilibrium

[N2O] = 1.68 moles/litre

[O2] = 2.91 + x = 2.91 + 1.68 = 4.59 moles/litre

[N2O3] = 2.91 - x = 2.91- 1.68 = 1.23 moles/litre

[N2O5] = 4.0 - 2.91 = 1.09 moles/litre

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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