Question

The equilibrium constant Kp for the formation of COBr2, is given below.
CO(g) + Br2(g) <-----> COBr2(g) Kp = 4.7 x 107 at 150°C

(a) Calculate Kc for the reaction as written above. (1 pt)
Answer ___________________

(b) Calculate Kp for the following reaction at 150°C. (Refer to the reaction and Kp = 4.7 x 107
from part a.) (1 pt)

3COBr2(g) <------> 3CO(g) + 3Br2(g)

Answer ___________________

(c) At a HIGHER temperature, a mixture of CO and Br2 is placed in a reaction flask with
concentrations [CO] = 0.0102 M and [Br2] = 0.00609 M. When the reaction comes to
equilibrium at the new temperature, [Br2] = 0.00301 M. Determine Kc at the higher
temperature. (1 pts)

CO(g) + Br2(g) <-------> COBr2(g)

Answer ___________________

(d) Based on your responses from Part (a) at 150°C compared ot Part (b) at the higher temp, does
the reaction appear to be endothermic or exothermic? Briefly justify your response. (1 pt)

Answer 1