Chapter 15: Problem 52
For the equilibrium $$ \mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \operatorname{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0 .\) If \(0.25 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) and \(0.55 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) are introduced into a 3.0-L container at \(400 \mathrm{~K},\) what will be the equilibrium concentrations of \(\mathrm{Br}_{2}, \mathrm{Cl}_{2}\), and \(\mathrm{BrCl}\) ?
Short Answer
Step by step solution
Write the Initial Concentrations
Define the Change in Concentrations
Write the Equilibrium Concentrations
Set Up the Equilibrium Expression
Solve for x
Identify the Valid x Value
Calculate the Equilibrium Concentrations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Constant
- If \(K_c\) is much greater than 1, the reaction favors the formation of products.
- If \(K_c\) is much less than 1, the reaction favors the formation of reactants.
Initial Concentration
- \(0.25 \mathrm{~mol}\) of \(\mathrm{Br}_2\)
- \(0.55 \mathrm{~mol}\) of \(\mathrm{Cl}_2\)
Knowing these helps in setting up the chemical reaction and determining how the reactants convert to products.
Quadratic Formula
The quadratic equation derived is:\[3x^2 - 0.9338x + 0.106813 = 0.\]Here, \(a = 3\), \(b = -0.9338\), \(c = 0.106813\). Solving this yields possible \(x\) values, representing reacted or produced concentrations. You choose the \(x\) that results in non-negative concentrations, keeping the solution physically meaningful.
Reaction Quotient
- If \(Q_c < K_c\), the reaction proceeds forward to reach equilibrium.
- If \(Q_c = K_c\), the reaction is already at equilibrium.
- If \(Q_c > K_c\), the reaction will shift backwards to reach equilibrium.
Le Chatelier's Principle
- Concentration: Adding or removing substances shifts the equilibrium to re-balance product and reactant concentrations.
- Pressure: For gaseous reactions, increasing pressure by reducing volume causes a shift towards the side with fewer moles of gas.
- Temperature: Raising temperature favors the endothermic direction, while lowering favors the exothermic.