Question

A(g) + 2B(g) → C(g) + D(g)

If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of K_p for the reaction at the temperature the reaction was run?

Answer 1

Set up the ICE chart for the reaction as below.

A (g) + 2 B (g) <======> C (g) + D (g)

Initial                                    1.00      1.00                       -            -

Change                                 -x         -2x                        +x         +x

Equilibrium                      (1.00 – x)(1.00 – 2x)                 x           x

It is given that the equilibrium pressure of C is P_C = 0.211 atm; therefore,

x = 0.211 atm.

Therefore, P_D = x = 0.211 atm.

P_A = (1.00 – 0.211) atm = 0.789 atm.

P_B = (1.00 – 2*0.211) atm = 0.578 atm.

The equilibrium constant for the reaction is given as

K_p = (P_C)(P_D)/(P_A)(2P_B)²

= (0.211 atm)(0.211 atm)/(0.789 atm)(0.578 atm)²

= 0.1689 atm^-1

≈ 0.169 atm^-1

The equilibrium constant for the reaction is 0.169 atm^-1 (ans, correct to 3 sig. figs).