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Chapter 15: Problem 79

For the equilibrium $$2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\operatorname{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(5.07 \mathrm{kPa}\) of IBr is placed in a 10.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

Short Answer

Expert verified
The partial pressures of all substances after the equilibrium is reached are: IBr: \(4.336\ kPa\), I2: \(0.367\ kPa\), and Br2: \(0.367\ kPa\).

Step by step solution

01

Calculate initial partial pressures of substances

Using the ideal gas law, we can find the initial number of moles of IBr and use it to determine the initial partial pressures of the substances. \(P_{IBr} = 5.07\ kPa\) \(V = 10.0\ L\) We are given the initial pressure directly: \(P_{initial} = P_{IBr, initial} = 5.07 \mathrm{kPa}\) For I2 and Br2, the initial pressure is 0, as they are not present at the beginning: \(P_{I_{2},initial} = P_{Br_{2},initial} = 0\) Now we have the initial partial pressures: - \(P_{IBr, initial} = 5.07\ kPa\) - \(P_{I_{2},initial} = 0\) - \(P_{Br_{2},initial} = 0\)
02

Set up an ICE table

Next, we will set up an ICE (Initial change equilibrium) table for the reaction: \[ \begin{array}{|c|c|c|c|} \hline & 2 \mathrm{IBr}(g) & \mathrm{I}_{2}(g) & \mathrm{Br}_{2}(g) \\ \hline \mathrm{Initial} & 5.07 kPa & 0 & 0 \\ \hline \mathrm{Change} & -2x & +x & +x \\ \hline \mathrm{Equilibrium} & 5.07 - 2x & x & x \\ \hline \end{array} \] Where x is the change in partial pressure of the substances as the reaction proceeds to equilibrium.
03

Use the equilibrium constant Kp to solve for x

We can now write an expression for the equilibrium expression using our ICE table values: \(K_p = \frac{P_{I_{2}} \cdot P_{Br_{2}}}{(P_{IBr})^2}\) Substitute the equilibrium partial pressures from the ICE table: \(8.5\times10^{-3} = \frac{x\cdot x}{(5.07-2x)^2}\) Now, we can solve this quadratic equation for x numerically or by trying some iterative approaches.
04

Calculating the equilibrium partial pressures

Upon solving the quadratic equation, we find that x = 0.367 kPa (only considering the physically meaningful positive value). Now we can plug this value of x back into our equilibrium row of our ICE table to find the equilibrium partial pressures for all the substances: - \(P_{IBr,equilibrium} = 5.07 - 2(0.367) = 4.336\ kPa\) - \(P_{I_{2},equilibrium} = 0.367\ kPa\) - \(P_{Br_{2},equilibrium} = 0.367\ kPa\) The partial pressures of all substances after the equilibrium is reached are: - IBr: \(4.336\ kPa\) - I2: \(0.367\ kPa\) - Br2: \(0.367\ kPa\)

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Most popular questions from this chapter

If \(K_{c}=1\) for the equilibrium $3 \mathrm{~A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$, what is the relationship between [A] and [B] at equilibrium?

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$. A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up $15.0 \mathrm{~L}\( of a solution that is initially \)0.275 \mathrm{M}$ in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

A 5.37 -g sample of \(\mathrm{SO}_{3}\) is placed in a 5.00-L container and heated to \(1000 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the total pressure in the container is 157 \(\mathrm{kPa}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Water molecules in the atmosphere can form hydrogenbonded dimers, \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSF.PR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The \(K_{p}\) for water dimer formation in the gas phase is 0.050 at $300 \mathrm{~K}\( and 0.020 at \)350 \mathrm{~K}$. Is water dimer formation endothermic or exothermic?

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: $2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)$. (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibriumconstant expression in terms of molarities for the reaction, using (solv) to indicate solvation.
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