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Chapter 15: Problem 14

Bromine and chlorine react to produce bromine monochloride according to the equation \\[ \mathrm{Br}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{BrCl}(\mathrm{g}) \\] \(0.2 \mathrm{mol}\) of bromine gas and \(0.2 \mathrm{mol}\) of chlorine gas are introduced into a sealed flask with a volume of \(5.0 \mathrm{dm}^{3}\). Under the conditions of the experiment, \(K=36.0 .\) How much BrCl will be present at equilibrium? (Section 15.4 )

Short Answer

Expert verified
0.34 mol of BrCl will be present at equilibrium.

Step by step solution

01

Write the Expression for Equilibrium Constant

The equilibrium constant expression for the reaction \( \mathrm{Br}_{2}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{BrCl}(\mathrm{g}) \) is given by:\[K = \frac{{[\mathrm{BrCl}]^2}}{{[\mathrm{Br}_2][\mathrm{Cl}_2]}}\]Where \([\mathrm{BrCl}], [\mathrm{Br}_2],\) and \([\mathrm{Cl}_2] \) are the equilibrium concentrations of BrCl, Br鈧�, and Cl鈧�, respectively.
02

Calculate Initial Concentrations

The initial moles of \( \mathrm{Br}_2 \) and \( \mathrm{Cl}_2 \) are both 0.2 mol, in a 5.0 dm鲁 flask:\[[\mathrm{Br}_2]_0 = \frac{0.2 \text{ mol}}{5.0 \text{ dm}^3} = 0.04 \text{ M}\]\[[\mathrm{Cl}_2]_0 = \frac{0.2 \text{ mol}}{5.0 \text{ dm}^3} = 0.04 \text{ M}\]
03

Set Up the Change in Concentration

Let \( x \) be the change in concentration for BrCl formed. The changes in concentration for the reactants will be \(-x\) for \( \mathrm{Br}_2 \) and \( \mathrm{Cl}_2 \), and \(+2x\) for \( \mathrm{BrCl} \) since the stoichiometry in the balanced equation gives 2 moles of \( \mathrm{BrCl} \) for every mol of \( \mathrm{Br}_2 \) and \( \mathrm{Cl}_2 \) that reacts.
04

Write Equilibrium Concentrations in Terms of x

At equilibrium, the concentrations are:\[[\mathrm{Br}_2] = 0.04 - x\]\[[\mathrm{Cl}_2] = 0.04 - x\]\[[\mathrm{BrCl}] = 2x\]
05

Substitute into Equilibrium Expression

Substituting into the equilibrium expression:\[K = \frac{(2x)^2}{(0.04 - x)(0.04 - x)}\]Given \( K = 36.0 \):\[36 = \frac{4x^2}{(0.04 - x)^2}\]
06

Solve for x

Let's solve the equation: \[4x^2 = 36(0.04 - x)^2\]Expanding:\[4x^2 = 36(0.0016 - 0.08x + x^2)\]\[4x^2 = 0.0576 - 2.88x + 36x^2\]Simplify and rearrange:\[0 = 32x^2 - 2.88x + 0.0576\]Solve this quadratic equation to find \(x\). Using the quadratic formula,\[x \approx 0.034\].
07

Determine Equilibrium Concentration of BrCl

Since \([\mathrm{BrCl}]= 2x\), substitute found \( x \):\[[\mathrm{BrCl}] = 2(0.034) = 0.068 \text{ M}\]Convert concentration to moles in the 5.0 dm鲁 flask:\[\text{Moles of } \mathrm{BrCl} = 0.068 \times 5.0 = 0.34 \text{ mol}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
In any reversible chemical reaction, the equilibrium constant, often symbolized as \( K \), plays a crucial role in determining the concentration of reactants and products at equilibrium.
  • The equilibrium constant is derived from the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of reactants raised to their stoichiometric coefficients.
  • For the reaction \( \mathrm{Br}_{2} + \mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{BrCl} \), the expression for \( K \) is given by \[ K = \frac{{[\mathrm{BrCl}]^2}}{{[\mathrm{Br}_2][\mathrm{Cl}_2]}}. \]
  • The value of \( K \), which is \( 36.0 \) in this case, helps us predict how far the reaction will proceed to produce \( \mathrm{BrCl} \).
A large \( K \) value means the reaction favors the formation of products, showing that at equilibrium there will be a higher concentration of \( \mathrm{BrCl} \). Understanding the equilibrium constant thus allows us to quantitatively describe chemical equilibria.
Concentration Calculation
Calculating concentrations is essential for setting up the problem and analyzing the equilibrium state.
  • Initially, we have 0.2 moles of \( \mathrm{Br}_2 \) and \( \mathrm{Cl}_2 \) each, introduced into a 5.0 dm鲁 flask. The initial concentration \( [\mathrm{Br}_2]_0 \) and \( [\mathrm{Cl}_2]_0 \) is calculated by dividing the moles by the volume of the flask:
  • \[ [\mathrm{Br}_2]_0 = [\mathrm{Cl}_2]_0 = \frac{0.2 \text{ mol}}{5.0 \text{ L}} = 0.04 \text{ M}. \]
  • This step sets the stage for the changes that occur as the reaction moves toward equilibrium.
Understanding how to calculate these initial conditions allows you to effectively predict the behavior of the system under study.
Reaction Stoichiometry
In chemical reactions, stoichiometry is vital for understanding how the amount of reactants relates to the formation of products.
  • The balanced reaction \( \mathrm{Br}_{2} + \mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{BrCl} \) indicates that one mole of \( \mathrm{Br}_{2} \) reacts with one mole of \( \mathrm{Cl}_{2} \) to produce two moles of \( \mathrm{BrCl} \).
  • When approaching equilibrium, a change \( x \) occurs where \( \mathrm{Br}_{2} \) and \( \mathrm{Cl}_{2} \) concentrations decrease by \( x \), while \( \mathrm{BrCl} \) increases by \( 2x \).
  • The stoichiometric coefficients guide us in setting up these change relationships as: \[ [\mathrm{Br}_2] = 0.04 - x, \]\[ [\mathrm{Cl}_2] = 0.04 - x, \]\[ [\mathrm{BrCl}] = 2x. \]
Understanding stoichiometry helps us predict how concentrations will change as the reaction proceeds.
Quadratic Equation
The analysis of equilibrium often leads to solving a quadratic equation, a mathematical tool that helps determine equilibrium concentrations.
  • We use the equilibrium expression and the known value of \( K \) to set up a quadratic equation as follows:
  • The relationship \[ K = \frac{(2x)^2}{(0.04 - x)^2} \] leads to \[ 4x^2 = 36(0.04 - x)^2. \]
  • Expanding and rearranging gives a quadratic equation: \[ 0 = 32x^2 - 2.88x + 0.0576. \]
  • The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is then used to find \( x \), the change in concentration.
Solving this equation gives \( x \approx 0.034 \), leading to the calculation of equilibrium amounts of \( \mathrm{BrCl} \). Mastering this technique is crucial for any chemistry student dealing with equilibria.

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Most popular questions from this chapter

\(\mathrm{CO}_{2}\) decomposes into \(\mathrm{CO}\) and \(\mathrm{O}_{2}\) over a platinum catalyst \\[ 2 \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \\] At 1 bar pressure, the fraction, \(\alpha,\) of \(\mathrm{CO}_{2}\) that reacts is 0.014 at \(1395 \mathrm{K}, 0.025\) at \(1443 \mathrm{K}\), and 0.047 at \(1498 \mathrm{K}\). (Section 15.5 ) (a) Calculate the equilibrium constant at \(1443 \mathrm{K}\) (b) Calculate \(\Delta_{r} H^{\theta}\) and \(\Delta_{r} S^{\theta}\) for the reaction.

Calculate the maximum quantity (in mol) of \(\mathrm{KIO}_{3}\) that can be added to \(250 \mathrm{cm}^{3}\) of a solution containing \(1.00 \times 10^{-3} \mathrm{mol} \mathrm{dm}^{-3}\) of \(\mathrm{Cu}^{2}\) ' (aq) without precipitating \(\mathrm{Cu}(\mathrm{IO} 3)_{2}(\mathrm{s}) . K_{\mathrm{sp}}=1.4 \times 10^{-7}\) for \(\mathrm{Cu}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s})\).

2.0 mol of carbon disulfide and \(4.0 \mathrm{mol}\) of chlorine react at constant temperature according to this equation \\[ \mathrm{CS}_{2}(\mathrm{g})+3 \mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{S}_{2} \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{CCl}_{4}(\mathrm{g}) \\] At equilibrium, \(0.30 \mathrm{mol}\) of tetrachloromethane are formed. How much of each of the other components is present in this equilibrium mixture? (Section 15.4 )

Nitrosy chloride (NOCI) decomposes to nitric oxide and chlorine when heated \\[ 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \\] In a mixture of all three gases at \(600 \mathrm{K}\), the partial pressure of NOCl is 0.88 bar, that of \(\mathrm{NO}\) is 0.06 bar, and the partial pressure of chlorine is 0.03 bar. At \(600 \mathrm{K}\), the equilibrium constant, \(K,\) is \(0.060 .\) (Section 15.2 and several others) (a) What is the value of the reaction quotient for this mixture? Is the mixture at equilibrium? (b) In which direction will the system move to reach equilibrium? (c) What will happen if an additional amount of NOCl(g) is injected into the reaction?

A student was investigating the following equilibrium reaction Which has an equilibrium constant of 0.220 at \(800^{\circ} \mathrm{C}\) \\[ \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \\] and did four experiments. (a) \(0.2 g\) of \(\mathrm{CaCO}_{3}(\mathrm{s})\) was heated to \(800^{\circ} \mathrm{C}\) in a \(1.0 \mathrm{dm}^{3}\) container (b) \(2.0 \mathrm{g}\) of \(\mathrm{CaCO}_{3}(\mathrm{s})\) was heated to \(800^{\circ} \mathrm{C}\) in a \(1.0 \mathrm{dm}^{3}\) container (c) \(0.2 g\) of \(\mathrm{CaCO}_{3}\) (s) was heated to \(800^{\circ} \mathrm{C}\) in a \(500 \mathrm{cm}^{3}\) container (d) \(2.0 g\) of \(\mathrm{CaCO}_{3}(\mathrm{s})\) was heated to \(800^{\circ} \mathrm{C}\) in a \(500 \mathrm{cm}^{3}\) container The pressure of \(\mathrm{CO}_{2}\) (g) measured in each case was ( 1 ) 0.18 bar, (ii) 0.22 bar, (iii) 0.22 bar, \((\text { iv) } 0.22\) bar. Explain these observations.
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