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Chapter 15: Problem 17

\(1.00 \times 10^{-3} \mathrm{mol} \mathrm{PCl}_{5}\) is introduced into a \(250.0 \mathrm{mL}\) flask, and equilibrium is established at \(284^{\circ} \mathrm{C}\) : \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) .\) The quantity of \(\mathrm{Cl}_{2}(\mathrm{g})\) present at equilibrium is found to be \(9.65 \times 10^{-4} \mathrm{mol}\) What is the value of \(K_{c}\) for the dissociation reaction at \(284^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The value of \(K_c\) for the dissociation reaction at \(284^{\circ}\mathrm{C}\) is approximately 107.04.

Step by step solution

01

Identify Initial Moles of Substances

From the problem, the initial moles of \(PCl_5\) is \(1.00 \times 10^{-3} \) mol. And it is given that the moles of \(Cl_2\) at equilibrium is \(9.65 \times 10^{-4}\) mol. Because the reaction has not started, there is initially no \(PCl_3\) and \(Cl_2\).
02

Determine the Moles at Equilibrium

At equilibrium, the moles of \(PCl_5\) used will be equal to the moles of \(Cl_2\) formed since they follow a 1:1 ratio based on the balanced chemical reaction. So the moles of \(PCl_5\) at equilibrium will be \(1.00 \times 10^{-3} - 9.65 \times 10^{-4} = 0.35 \times 10^{-4}\) mol.
03

Calculate the Concentrations at Equilibrium

To calculate the concentration, we need to divide the moles by the volume of the flask. The volume of the flask is 250.0 mL which is equivalent to 0.25 L. So, the concentration of \(PCl_5\) is \(0.35 \times 10^{-4} / 0.25 = 1.4 \times 10^{-4}\) M, the concentration of \(Cl_2\) is \(9.65 \times 10^{-4} / 0.25 = 3.86 \times 10^{-3}\) M, and the concentration of \(PCl_3\) is equal to the concentration of \(Cl_2\) which is \(3.86 \times 10^{-3}\) M.
04

Calculate the Equilibrium Constant (\(K_c\))

Using the equilibrium constant expression for the balanced chemical reaction \(K_c = [PCl_3][Cl_2]/[PCl_5]\), we substitute the concentrations into the expression, then \(K_c = (3.86 \times 10^{-3})(3.86 \times 10^{-3})/(1.4 \times 10^{-4})\).

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Most popular questions from this chapter

Determine \(K_{c}\) for the reaction $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NOBr}(\mathrm{g})$$ from the following information (at \(298 \mathrm{K}\) ). $$\begin{aligned} 2 \mathrm{NO}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) K_{\mathrm{c}}=2.1 \times 10^{30} \\ \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NOBr}(\mathrm{g}) \quad K_{\mathrm{c}}=1.4 \end{aligned}$$

For the reaction $$ \mathrm{A}(\mathrm{s}) \rightleftharpoons \mathrm{B}(\mathrm{s})+2 \mathrm{C}(\mathrm{g})+\frac{1}{2} \mathrm{D}(\mathrm{g}) \quad \Delta H^{\circ}=0 $$ (a) Will \(K_{p}\) increase, decrease, or remain constant with temperature? Explain. (b) If a constant-volume mixture at equilibrium at 298 K is heated to 400 K and equilibrium re-established, will the number of moles of \(\mathrm{D}(\mathrm{g})\) increase, decrease, or remain constant? Explain.

Lead metal is added to \(0.100 \mathrm{M} \mathrm{Cr}^{3+}(\mathrm{aq}) .\) What are \(\left[\mathrm{Pb}^{2+}\right],\left[\mathrm{Cr}^{2+}\right],\) and \(\left[\mathrm{Cr}^{3+}\right]\) when equilibrium is established in the reaction? $$\begin{aligned} \mathrm{Pb}(\mathrm{s})+2 \mathrm{Cr}^{3+}(\mathrm{aq}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cr}^{2+}(\mathrm{aq}) & \\ K_{\mathrm{c}}=3.2 \times 10^{-10} & \end{aligned}$$

At \(2000 \mathrm{K}, K_{c}=0.154\) for the reaction \(2 \mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If a \(1.00 \mathrm{L}\) equilibrium mixture at \(2000 \mathrm{K}\) contains \(0.10 \mathrm{mol}\) each of \(\mathrm{CH}_{4}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (a) what is the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) present? (b) Is the conversion of \(\mathrm{CH}_{4}(\mathrm{g})\) to \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) favored at high or low pressures? (c) If the equilibrium mixture at \(2000 \mathrm{K}\) is transferred from a 1.00 L flask to a 2.00 L flask, will the number of moles of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) increase, decrease, or remain unchanged?

The standard enthalpy of reaction for the decomposition of calcium carbonate is \(\Delta H^{\circ}=813.5 \mathrm{kJmol}^{-1}\) As temperature increases, does the concentration of calcium carbonate increase, decrease, or remain the same? Explain.
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