Question

Consider the following reaction and its corresponding equilibrium constant to answer the next two questions:

                        P₄ (g) + 6 H₂ (g) Û 4 PH₃ (g)        K_p = 6.48 x 10^-6 (at 25 ^oC)

17.       What is the value of K_c for the following reaction given that K_p equals 6.48 x 10^-6 (at 25 ^oC)?

            a) K_c = 3.98 x 10^-12                           b) K_c = 4.42 x 10^-10                          c) K_c = 5.60 x 10^-5

            d) K_c = 9.49 x 10^-2                           e) K_c = 9.848 x 10⁴

18.       Given the reaction above and it’s corresponding equilibrium constant, what is the value of the equilibrium constant defined below?

                        2 PH₃ (g) Û ½ P₄ (g) + 3 H₂ (g)                 K_net = ??

            a) K_net = 4.20 x 10^-11             b) K_net = 2.55 x 10^-3                          c) K_net = 3.93 x 10²

            d) K_net = 1.54 x 10⁵                          e) K_net = 2.38 x 10¹⁰

^{I understand how to get number 17. I need help with 18.}

Answer 1

17.
P4 (g) + 6 H2 (g) <---> 4 PH3 (g) Kp = 6.48 x 10-6 (at 25 oC)

Kp = Kc*(RT)^Dn

Kp = 6.48*10^-6

Dn = change in no of moles = 4 - (6+1) = -3

R = 0.0821 L.atm.K-1.mol-1

T = 25 c = 298 k

6.48*10^-6 = x*(0.0821*298)^(-3)

x = Kc = 0.095

Kc = 9.5*10^-2

answer: d

18.     2 PH3 (g) <---> ½ P4 (g) + 3 H2 (g)

   equation is reversed and divided with 2

so that,

   Knet = (1/Kp)^(1/2)

         = (1/(6.48*10^-6))^(1/2)

         = 392.8

         = 3.93*10^2

answer: c