We are being asked to calculate the equilibrium CoBr_{2} of the three gases in the following reaction:

COBr_{2}(g) ⇌ CO(g) + Br_{2}(g) K_{c} = 0.190

**When dealing with equilibrium and K:**

• **K **→ equilibrium units are in molarity

• **K** is an equilibrium expression:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}$

▪ **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 COBr_{2}.**Step 4:** Calculate the concentration of COBr_{2} when the volume of the flask was halved

1.00 moles of COBr_{2} were placed in a 2.00 liter flask and heated to 73 °C. At that temperature, K_{c} = 0.190 for the reaction COBr_{2}(g) ⇌ CO(g) + Br_{2}(g).

Part I: Find the equilibrium concentration of COBr_{2} (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 COBr_{2} (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