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Chapter 15: Problem 80

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 -L. vessel at \(60^{\circ} \mathrm{C}\) , the vessel is charged with 0.0500 \(\mathrm{g}\) of \(\mathrm{BCl}_{3}(g) .\) What is the equilibrium concentration of \(\mathrm{PH}_{3} ?\)

Short Answer

Expert verified
The equilibrium concentration of PH3 is approximately 0.00107 M.

Step by step solution

01

Convert Kp to Kc

To convert Kp to Kc, we'll use the relationship between Kp and Kc: \[K_{p} = K_{c}(RT)^{\Delta n}\] where R is the ideal gas constant (0.08206 L⋅atm/mol⋅K), T is the temperature in Kelvin, and Δn is the difference in moles of gaseous products and reactants. In this case: - Δn = (1 mole of PH3 + 1 mole of BCl3) - 0 moles of PH3BCl3 = 2 - 0 = 2 - T = 60°C + 273.15 = 333.15 K - Kp = 0.052 Substitute these values into the equation and solve for Kc: \[0.052 = K_{c}(0.08206 \times 333.15)^{2}\] \[K_{c} = \frac{0.052}{(0.08206 \times 333.15)^{2}}\] Now calculate the value of Kc:
02

Calculate Kc

By calculating the expression from Step 1, we get: \[K_{c} \approx 8.71 \times 10^{-6}\]
03

Set up an ICE table

To find the equilibrium concentration of PH3, we first set up an ICE (Initial, Change, Equilibrium) table: | | PH3(g) | BCl3(g) | PH3BCl3(s) | |-------|--------|---------|------------| | Initial | 0 | 0.0500/1.5 | 3/1.5 | | Change | +x | +x | -x | | Equilibrium | x | 0.0500/1.5 + x | 3/1.5 - x |
04

Write the expression for Kc

Now, we'll write the Kc expression for the reaction: \[K_{c} = \frac{[\mathrm{PH}_{3}][\mathrm{BCl}_{3}]}{[\mathrm{PH}_{3}\mathrm{BCl}_{3}]}\] Substitute the equilibrium concentrations from the ICE table: \[8.71 \times 10^{-6} = \frac{x(x + 0.0500/1.5)}{3/1.5 - x}\] Solve for x, which is the equilibrium concentration of PH3.
05

Calculate the equilibrium concentration of PH3

After solving the equation from Step 4, we find: \[x \approx 0.00107\, M\] Therefore, the equilibrium concentration of PH3 is approximately 0.00107 M.

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Most popular questions from this chapter

Calculate \(K_{c}\) at 303 \(\mathrm{K}\) for \(\mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) if \(K_{p}=34.5\) at this temperature.

The water-gas shift reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) is used industrially to produce hydrogen. The reaction enthalpy is \(\Delta H^{\circ}=-41 \mathrm{kJ}\) . (a) To increase the equilibrium yield of hydrogen would you use high or low temperature? ( b) Could you increase the equilibrium yield of hydrogen by controlling the pressure of this reaction? If so would high or low pressure favor formation of \(\mathrm{H}_{2}(g) ?\)

Consider the hypothetical reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{B}(g) . \mathrm{A}\) flask is charged with 0.75 atm of pure \(\mathrm{A},\) after which it is allowed to reach equilibrium at \(0^{\circ} \mathrm{C}\) . At equilibrium, the partial pressure of \(\mathrm{A}\) is 0.36 atm. (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?(\mathbf{c})\) What could we do to maximize the yield of B?

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at 1000 \(\mathrm{K}\) . (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2},\) or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) (c) Calculate \(K_{c}\) for \(\operatorname{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\)

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?
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