Question

Equal amounts of carbon monoxide and bromine react at 70° C to form carbonyl bromide. CO (g) + Br2 (g) <-->COBr2 (g)

a) What is the equilibrium constant Kp at this temperature of the reaction when there is 47.2 volume-% of carbon monoxide in the equilibrium mixture and the total pressure in the equilibrium mixture is 1.0 MPa? Give the answer to four decimal places. The balance equation unit does not need to be written into the answer. (use partial pressure)

b)

b) How many percent of bromine has reacted?

c) How many percent of bromine reacts if the molar ratio of CO and Br2 is 6: 1?

Answer 1

Part (a)

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

As it is given that 47.2 is the volume percent of the CO, hence the volume % of the Bromine will also be the same as they both are equimolar reactants in this reaction.

47.2 + 47.2 = 94.4

Hence, the remaining 5.6 % of volume will be occupied by Carbonyl Bromide.

Now let us calculate the Kp.

$K_p = \frac{5.6}{47.2*47.2}$

Hence, this is the value of our Kp.

Part (B)

Initially, Bromine was 50 Volume % (Equimolar Bromine and CO reactants).

At equilibrium, Bromine was 47.2 Volume % hence 2.8 Volume % of Bromine has reacted.

Hence Out of Total Bromine 5.6 Percent of Bromine has reacted.

Part (C)

The questions asks that how many percent of bromine reacts if the molar ratio of CO and Br2 is 6: 1.

As the 1 mole of CO reacts with 1 mole of Bromine, if more moles of CO are added they will be left unreacted as no moles of Bromine will be present to react with them. Hence all the Bromine will be reacted. Hence 100 % of Bromine will be reacted.