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Chapter 12: Problem 36

At a certain temperature, \(K\) is \(0.040\) for the decomposition of two moles of bromine chloride gas (BrCl) to its elements. An equilibrium mixture at this temperature contains bromine and chlorine gases at equal partial pressures of \(0.0493\) atm. What is the equilibrium partial pressure of bromine chloride?

Short Answer

Expert verified
Answer: The equilibrium partial pressure of bromine chloride (BrCl) is approximately 0.0547 atm.

Step by step solution

01

Write the balanced chemical equation

For the decomposition of bromine chloride into its elements, we have the balanced chemical equation: \(2BrCl_{(g)} \rightleftharpoons Br_{2(g)} + Cl_{2(g)}\)
02

Set up an ICE table

We will set up an ICE table to keep track of the initial and equilibrium partial pressures of the substances in the reaction. Let P be the partial pressure of BrCl at equilibrium. | Reaction | Initial | Change | Equilibrium | |:--------------:|:-------:|:--------:|:-----------:| | 2BrCl(g) | - | +2x | P | | Br2(g) | 0 | - x | 0.0493 | | Cl2(g) | 0 | - x | 0.0493 | Given that the equilibrium partial pressures of bromine and chlorine gases are equal, both the change in the partial pressure of Br2 and Cl2 is x.
03

Write the equilibrium constant expression

The equilibrium constant expression for the reaction is given by: \(K = \frac{[Br_2][Cl_2]}{[BrCl]^2}\) In terms of partial pressures, the expression becomes: \(K = \frac{P(Br_2)P(Cl_2)}{P^2(BrCl)}\)
04

Substitute the given value of K and partial pressures

We are given K = 0.040, and the equilibrium partial pressures of bromine and chlorine gases are 0.0493 atm. Plug these values into the equilibrium constant expression: \(0.040 = \frac{(0.0493)(0.0493)}{P^2}\)
05

Solve for the equilibrium partial pressure of bromine chloride, P

Rearrange the equation to solve for the partial pressure of BrCl (P): \(P^2=\frac{(0.0493)(0.0493)}{0.040}\) Now, take the square root of both sides: \(P=\sqrt{\frac{(0.0493)(0.0493)}{0.040}}\) Calculate the result: \(P\approx 0.0547 \,\text{atm}\)
06

State the final answer

The equilibrium partial pressure of bromine chloride (BrCl) is approximately 0.0547 atm.

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Most popular questions from this chapter

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?

Consider the system $$\mathrm{A}(g)+2 \mathrm{~B}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{D}(g)$$ at \(25^{\circ} \mathrm{C}\). At zero time, only \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) are present. The reaction reaches equilibrium \(10 \mathrm{~min}\) after the reaction is initiated. Partial pressures of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{D}\) are written as \(P_{\mathrm{A}}, P_{\mathrm{B}}\), and \(P_{\mathrm{D}}\). Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required). (a) \(P_{\mathrm{D}}\) at \(11 \mathrm{~min}\) ________ \(P_{\mathrm{D}}\) at \(12 \mathrm{~min} .\) (b) \(P_{\mathrm{A}}\) at \(5 \mathrm{~min}\) \(P_{\mathrm{A}}\) ______ at \(7 \mathrm{~min}\) (c) \(K\) for the forward reaction ______ \(K\) for the reverse reaction. (d) At equilibrium, \(K\)______Q. (e) After the system is at equilibrium, more of gas \(\mathrm{B}\) is added. After the system returns to equilibrium, \(K\) before the addition of \(B\) \(K\) _____ after the addition of \(\mathrm{B}\). (f) The same reaction is initiated, this time with a catalyst. \(K\) for the system without a catalyst _____ \(K\) for the system with a catalyst. (g) \(K\) for the formation of one mole of \(\mathrm{D}\) \(K\) _____ for the formation of two moles of \(\mathrm{D}\). (h) The temperature of the system is increased to \(35^{\circ} \mathrm{C} . P_{\mathrm{B}}\) at equilibrium at \(25^{\circ} \mathrm{C} \longrightarrow P_{\mathrm{B}}\) _______at equilibrium at \(35^{\circ} \mathrm{C}\). (i) Ten more grams of \(\mathrm{C}\) are added to the system. \(P_{\mathrm{B}}\) before the addition of \(\mathrm{C} \quad P_{\mathrm{B}}\) _____ after the addition of \(\mathrm{C}\).

At \(25^{\circ} \mathrm{C}, K=2.2 \times 10^{-3}\) for the reaction $$\mathrm{ICl}(g) \rightleftharpoons \frac{1}{2} \mathrm{I}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ Calculate \(K\) at \(25^{\circ} \mathrm{C}\) for (a) the decomposition of ICl into one mole of iodine and chlorine. (b) the formation of two moles of \(\operatorname{ICl}(g)\).

At \(460^{\circ} \mathrm{C}\), the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g)$$ has \(K=84.7\). All gases are at an initial pressure of \(1.25\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{1}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{P_{\mathrm{IF}}}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{C}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\) (d) \(K=\frac{\left(P_{\mathrm{NO}}\right)^{2}\left(P_{\mathrm{H}_{3} \mathrm{O}}\right)^{4}\left[\mathrm{Cu}^{2+}\right]^{3}}{\left[\mathrm{NO}_{3}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{8}}\)
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