Chapter 16: Problem 17
Carbonyl bromide decomposes to carbon monoxide and bromine. $$ \operatorname{COBr}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ \(K\) is 0.190 at \(73^{\circ} \mathrm{C} .\) If you place \(0.500 \mathrm{mol}\) of \(\mathrm{COBr}_{2}\) in a 2.00-L flask and heat it to \(73^{\circ}\) C, what are the equilibrium concentrations of \(\mathrm{COBr}_{2}, \mathrm{CO},\) and \(\mathrm{Br}_{2} ?\) What percentage of the original COBr, decomposed at this temperature?
Short Answer
Step by step solution
Initial Concentration Calculation
Define Change in Concentration
Write the Equilibrium Expression
Solve for x
Calculate Equilibrium Concentrations
Calculate Percentage Decomposed
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Constant
- If \( K \) > 1, the products are favored.
- If \( K \) < 1, the reactants are favored.
- If \( K \) = 1, neither side is favored predominantly.
ICE Table
- Initial: Concentrations of all substances at the start of the reaction.
- Change: The change in concentration as the system reaches equilibrium, often expressed in terms of \( x \), a variable that represents the amount that reacts or is formed.
- Equilibrium: The final concentrations of substances when the reaction is at equilibrium.
Concentration Calculation
- \([\text{CO}] = x = 0.036\, \text{M}\)
- \([\text{Br}_2] = x = 0.036\, \text{M}\)
- \([\text{COBr}_2] = 0.250 - x = 0.214\, \text{M}\)