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Chapter 15: Problem 78

For the equilibrium $$ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) $$ \(K_{p}=0.052\) at \(60^{\circ} \mathrm{C}\). (a) Calculate \(K_{c}\) (b) After \(3.00 \mathrm{~g}\) of solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed \(1.500\) - \(\mathrm{L}\) vessel at \(60{ }^{\circ} \mathrm{C}\), the vessel is charged with \(0.0500 \mathrm{~g}\) of \(\mathrm{BCl}_{3}(\mathrm{~g})\). What is the equilibrium concentration of \(\mathrm{PH}_{3}\) ?

Short Answer

Expert verified
The equilibrium concentration of PH3 is approximately 1.03 x 10^{-4} M.

Step by step solution

01

Write the expression for Kp in terms of Kc

Recall that Kp and Kc are related by the equation: \[K_p = K_c(RT)^{\Delta n}\] where R is the gas constant (0.0821 Latm/molK), T is the temperature in Kelvin, and Δn is the difference in the moles of the gaseous products and reactants.
02

Determine the value of Δn

In the balanced equilibrium reaction, \[\mathrm{PH}_{3}\mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g)\] Δn = moles of gaseous products - moles of gaseous reactants = (1 + 1) - 0 = 2.
03

Convert the temperature to Kelvin

Given that the temperature is 60°C, we will first need to convert it to Kelvin: \[60^{\circ} \mathrm{C} = 60 + 273.15 = 333.15 \mathrm{K}\]
04

Calculate Kc

Now use the given value of Kp and the calculated value of Δn to find Kc: \[K_c = \frac{K_p}{(RT)^{\Delta n}}\] \[K_c = \frac{0.052}{(0.0821 \times 333.15)^2}\] \[K_c \approx 1.87 \times 10^{-6}\] (b) Finding the equilibrium concentration of PH3
05

Calculate the moles of the reactant and product

Given that 3.00 g of solid PH3BCl3 is added, we need to convert it to moles: \[\text{moles of PH}_{3}\text{BCl}_{3} = \frac{3.00 \text{ g}}{269.65 \text{ g/mol}} \approx 0.0111 \text{ mol}\] Similarly, as 0.0500 g of BCl3(g) is already present in the vessel, we need to convert it to moles: \[\text{moles of BCl}_{3(g)} = \frac{0.0500 \text{ g}}{117.16 \text{ g/mol}} \approx 4.27 \times 10^{-4}\text{ mol}\]
06

Set up an ICE table

Initially, we have 0 mol of PH3 in the reaction mixture. Now, let's set up an ICE table (Initial, Change, Equilibrium) to find the equilibrium concentrations of the reactants and products. | | PH3BCl3(s) | PH3(g) | BCl3(g) | |---|-----------------|---------------------|---------------------| | I | 0.0111 mol | 0 mol | 4.27 x 10^-4 mol | | C | -x mol | +x mol | +x mol | | E | 0.0111-x mol | x mol | (4.27 x 10^-4+x) mol | Where x is the mol change of PH3 and BCl3.
07

Write the expression for Kc

Using the equilibrium concentrations from the ICE table, the expression for Kc becomes: \[K_c = \frac{[\mathrm{PH}_3][\mathrm{BCl}_3]}{[\mathrm{PH}_3\mathrm{BCl}_3]}\]
08

Calculate the value of x

Plug the equilibrium concentrations from the ICE table into the Kc expression: \[1.87 \times 10^{-6} = \frac{x(4.27 \times 10^{-4} + x)}{0.0111 - x}\] This is a quadratic equation in x. Solve the equation for x (mol of PH3) using the quadratic formula, substituting the given values.
09

Determine the equilibrium concentration of PH3

Once you have solved for x, you can use that to find the equilibrium concentration of PH3: \[[\mathrm{PH}_3] = \frac{x \text{ mol}}{1.500 \text{ L}}\] Calculate the value to obtain the equilibrium concentration of PH3.

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Most popular questions from this chapter

Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C},\) where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .\) At equilibrium it is found that \([\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{I}_{2}\right]=4.79 \times 10^{-4} \mathrm{M} .\) What is the value of \(K_{c}\) at this temperature?

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; K_{p}=5.0 \times 10^{12}\) (b) \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) ; K_{c}=5.8 \times 10^{-18}\)

For the equilibrium $$2 \mathrm{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) 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?

Explain why we normally exclude pure solids and liquids from equilibrium- constant expressions.

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and \(\mathrm{HI}\) at \(458{ }^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{~mol} \mathrm{H}_{2}, 0.112 \mathrm{~mol} \mathrm{I}_{2},\) and \(0.775 \mathrm{~mol} \mathrm{HI}\) in a \(5.00-\mathrm{L}\) vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 mol of HI?
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