Question

1. 1.00 mole of gaseous methane was put into a 1.0L flask ad 483K. The equilibrium mixture as depicted below was found. The equilibrium constant, K_C, is

1.50 moles Moles of reactants and products 0.75 moles -0.25 moles 0.204/ CHOH 0.20 CH OH Time

CO(g) + 2H₂(g) $\rightleftharpoons$ CH₃OH(g)

a) will be greater than 1 since products are favored

b) will be less than 1 since reactants are favored

c) neither statement above is correct because there are equal amounts of reactants and products

2. For the reaction given below, the diagram represents an equilibrium mixture. Calculate the pressure-based equilibrium constant, K_p, at this temperature. Assume that each molecule represents 1atm of pressure.

a) 3

b) 6

c) 12

d) 26

e) 108

3. The following equilibrium has a K = 23.2 at 600K.

H₂O(g) + CO(g)$\rightleftharpoons$ H₂(g) + CO₂(g)

What is the value of K_p at this temperature?

a) 554

b) 0.0101

c) 0.483

d) 1110

e) 23.2

4. The following equilibrium has a K = 3.0 at 298K.

2NO(g) + Br₂(g) $\rightleftharpoons$ 2NOBr(g)

What is the value of K_p at this temperature?

a) 3.0

b) 0.12

c) 073.6

d) 1810

e) 0.00500

Answer 1

I) Co + 2 H₂ CH₃OH Kc = [CH3OH] [Co] [Hz]2 0.25 0.75 (1.50)2 - b) will be less than I since reactants are favored. 2 NO 2) - 2 NO2 = 2 NO + O2 3 atm 6 atm 3 atm 1. Kp = Pro por = 6 x 3 = 12 6 þ ñoz 3) Kp = Kc LRT jong • H, () + CO3) = H3 + CO, (3) A4 = 2-2 = 0 =Kp = Kc = 23.2 e) 23.2 2 NO(g) + Brz (g) = 2 NOBY (9) ; Ana = 2-(1+3) --|| Kp = Ke (RT) Ang = 3.0 (0.0821 X298).' → Kp = 0.12 670.12