Problem: 1.00 moles of COBr2 were placed in a 2.00 liter flask and heated to 73 °C. At that temperature, Kc = 0.190 for the reaction COBr2(g) ⇌ CO(g) + Br2(g).Part I: Find the equilibrium concentration of COBr2 (in M).Part II: After equilibrium was established, the volume of the flask was halved (to 1.00 liters). Find the new equilibrium concentration of COBr2 (in M).Answers are for Part I; Part II (in M)A. 0.272 ; 0.648B. 0.648; 0.324C. 0.040; 0.080D. 0.040; 0.020E. 0.272; 0.544

FREE Expert Solution

We are being asked to calculate the equilibrium CoBr2 of the three gases in the following reaction:

 COBr2(g) ⇌ CO(g) + Br2(g)                Kc = 0.190


When dealing with equilibrium and K:

K → equilibrium units are in molarity 
K is an equilibrium expression:

K=productsreactants

only aqueous and gaseous species are included in the equilibrium expression
▪ the coefficient of each compound in the reaction equation will be the exponent of the concentrations in the equilibrium expression


We're going to calculate for the equilibrium concentrations using the following steps:

Step 1: Construct an ICE chart for the equilibrium reaction.
Step 2: Calculate the change (x) in the reaction.
Step 3: Calculate the equilibrium concentration of COBr2.
Step 4: Calculate the concentration of COBr2 when the volume of the flask was halved

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Problem Details

1.00 moles of COBr2 were placed in a 2.00 liter flask and heated to 73 °C. At that temperature, Kc = 0.190 for the reaction COBr2(g) ⇌ CO(g) + Br2(g).

Part I: Find the equilibrium concentration of COBr2 (in M).

Part II: After equilibrium was established, the volume of the flask was halved (to 1.00 liters). Find the new equilibrium concentration of COBr2 (in M).

Answers are for Part I; Part II (in M)

A. 0.272 ; 0.648

B. 0.648; 0.324

C. 0.040; 0.080

D. 0.040; 0.020

E. 0.272; 0.544