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Chapter 15: Problem 79

For the equilibrium $$2 \operatorname{IBr}(g) \Longrightarrow \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C} .\) If 0.025 atm of IBr is placed in a 2.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

Short Answer

Expert verified
At equilibrium with an initial pressure of 0.025 atm for IBr, the partial pressures of IBr, I2, and Br2 are approximately 0.0164 atm, 0.00431 atm, and 0.00431 atm, respectively.

Step by step solution

01

Write down the balanced reaction and set up the ICE table.

The balanced reaction is given in the exercise: $$2 \operatorname{IBr}(g) \Longrightarrow \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ We set up an ICE table to keep track of the initial pressures, changes in pressures, and final equilibrium pressures. | | IBr | ⇌ | I2 | + | Br2 | |--------|-----|----|-----|---|------| |Initial | 0.025 atm | | 0 atm | | 0 atm | |Change | -2x | | +x | | +x | |Equilibrium | 0.025-2x | | x | | x | We know that the total pressure of IBr at the beginning is 0.025 atm. Since the reaction is at equilibrium, we need to find x.
02

Write the equation for Kp and substitute the equilibrium pressures.

The equation for Kp is given as: $$K_{p}=\frac{[\mathrm{I}_{2}] [\mathrm{Br}_{2}]}{[\operatorname{IBr}]^2}$$ Since we are dealing with partial pressures, we can just substitute the equilibrium pressures (ICE table) into the equation: $$8.5 \times 10^{-3}=\frac{(x)(x)}{(0.025-2x)^2}$$
03

Solve the equation for x.

To find the value of x, we need to solve this equation. It is a quadratic equation, which can be simplified as follows: $$8.5 \times 10^{-3} = \frac{x^2}{(0.025-2x)^2}$$ Rearrange the equation: $$x^2 = 8.5 \times 10^{-3}(0.025 - 2x)^2$$ Expand and solve for x: $$x^2 = 8.5 \times 10^{-3}(4x^2 - 0.1x + 0.000625)$$ $$x^2=34x^4 - 8.5x^3 + 5.31 \times 10^{-6}x^2$$ $$x^2 - 34x^4 + 8.5x^3 - 5.31 \times 10^{-6}x^2 = 0$$ Solving this equation for x, we find two possible values, one positive and one negative. Since x represents changes in pressure, the negative value doesn't make sense in this context so we only consider the positive value: $$x \approx 0.00431 $$
04

Determine the partial pressures of all substances at equilibrium.

With the value of x, we can now substitute it back into the ICE table and find the partial pressures of all substances at equilibrium: | | IBr | ⇌ | I2 | + | Br2 | |--------|-----|----|-----|---|------| |Initial | 0.025 atm | | 0 atm | | 0 atm | |Change | -2x | | +x | | +x | |Equilibrium | 0.025-2x | | x | | x | The equilibrium partial pressures are: IBr: 0.025 - 2(0.00431) = 0.0164 atm I2: 0.00431 atm Br2: 0.00431 atm So at equilibrium, the partial pressures of IBr, I2, and Br2 are approximately 0.0164 atm, 0.00431 atm, and 0.00431 atm, respectively.

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Most popular questions from this chapter

A \(0.831-\) g sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 -L container and heated to 1100 \(\mathrm{K}\) . The 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 1.300 atm. Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at 1100 \(\mathrm{K}\) .

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

At \(800 \mathrm{K},\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5} .\) If an equilibrium mixture in a 10.0 -L. vessel contains \(2.67 \times 10^{-2} \mathrm{g}\) of I(g), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24^{\circ} \mathrm{C} .\) The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \Longrightarrow \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ At equilibrium, the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) taken together) is 0.614 atm. What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

Consider the reaction $$\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)$$ At \(25^{\circ} \mathrm{C},\) the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess CaSO \(_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4},\) what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{L},\) what is the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?
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