Question

9. The decomposition of N₂O₅is first order with a rate constant of 4.80 x 10^-4 s^-1at 45 ºC. If the initial concentration is

1.65 x 10^-2 M, what is the concentration after 825 s? What is the half life? How long would it take for the concentration of N₂O₅ to decrease to 1.00 x 10^-2 M from its initial value?

10. The following reaction has a K_c = 3.92 at 1200 Kelvin:     CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)

At equilibrium, the reaction vessel contains 0.30 M CO , 0.10 M H₂ , and 0.020 M H₂O

Write the equilibrium constant expression. What is the [CH₄] at equilibrium? Calculate the K_p (for atm).

answers:

9. 0.0111 M, 24.1 min, 17.4 min

10. K_c = [CH₄][H₂O]         [CH₄] = 0.059 M     K_p = 4.04 x 10^-4

               [CO][H₂]

Answer 1

9)

k = 4.80 x 10^-4 s^-1 at 45 ºC.

Ao = initial concentration = 1.65 x 10^-2 M,

part 1 )

half-life = 0.693 / k

          = 0.693 / 4.8 x 10^-4

            = 1443.75 sec

            = 24.1 min

part 2 )

t = 825 s

At = ?

k = 1/t * ln (Ao /At)

4.8 x 10^-4 = 1/ 825 * ln (1.65 x 10^-2 / At)

At = 0.0111 M

after 825 sec concentration = 0.0111 M

part 3 )

t = 1/k * ln (Ao /At)

t = 1/ 4.8 x 10^-4 * ln (1.65 x 10^-2 / 1.0 x 10^-2)

t = 1043.88 sec = 17.4 min

17.4 sec will take for the concentration of N₂O₅ to decrease to 1.00 x 10^-2 M from its initial value

10 )

   CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)

equilibrium constant expression. = Kc = [CH4][H2O]/[CO][H2]^3

3.93 = [CH4] x 0.02 / 0.30 x 0.1^3

[CH4] = 0.059 M

Kp = Kc (RT)Dn

Kp = 3.93 x (0.0821 x 1200)^-2

Kp = 4.04 x 0^-4