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Chapter 13: Problem 80

For the following reaction at a certain temperature $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ it is found that the equilibrium concentrations in a \(5.00\) -L rigid container are \(\left[\mathrm{H}_{2}\right]=0.0500 M,\left[\mathrm{~F}_{2}\right]=0.0100 M\), and \([\mathrm{HF}]=\) \(0.400 \mathrm{M}\). If \(0.200\) mole of \(\mathrm{F}_{2}\) is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Short Answer

Expert verified
The new equilibrium concentrations after the addition of $0.200$ mole of $\mathrm{F}_{2}$ are approximately: \([H_2] = 0.2712 M\), \([F_2] = 0.2288 M\), and \([HF] = 1.9576 M\).

Step by step solution

01

Determine the Initial Moles

Before the addition of F₂, the equilibrium concentrations are given as: \([H_2] = 0.0500 M\) \([F_2] = 0.0100 M\) \([HF] = 0.400 M\) The volume of the container is 5.00 L. To find the moles of each species, we can multiply the molarity by the volume: Moles of \(H_2 = 0.0500 M \times 5.00 L = 0.250\, mol\) Moles of \(F_2 = 0.0100 M \times 5.00 L = 0.050\, mol\) Moles of \(HF = 0.400 M \times 5.00 L = 2.000\, mol\)
02

Determine the Initial Moles After Addition of F₂

We need to add 0.200 mol of F₂ to the mixture. Initial moles of \(H_2 = 0.250\, mol\) Initial moles of \(F_2 = 0.050\, mol + 0.200\, mol = 0.250\, mol\) Initial moles of \(HF = 2.000\, mol\)
03

Set up the ICE Table

An ICE (Initial, Change, Equilibrium) table helps determine the changes in the moles and concentrations of each species as the reaction reaches a new equilibrium. ``` | H₂ | F₂ | HF --------------------------- Initial | 0.25 | 0.25 | 2.00 Change | +x | -x | -2x Final |0.25+x|0.25-x|2.0-2x ```
04

Calculate the Reaction Quotient (Q) and Equilibrium Constant (K)

From the initial equilibrium concentrations, we can calculate the reaction quotient, Q, and the equilibrium constant, K. \(Q = \frac{[HF]^2}{[H_2][F_2]} = \frac{(0.400)^2}{(0.0500)(0.0100)} = 3.20\) Since the system is already at equilibrium, this is also the value of the equilibrium constant, K: \(K = Q = 3.20\)
05

Solve for x Using the K Expression and the ICE Table

Using the equilibrium constant, K, and the final concentrations from the ICE table, we can solve for x. \(K = \frac{[HF]^2}{[H_2][F_2]} = \frac{(2.0-2x)^2}{(0.25+x)(0.25-x)}\) Substitute K = 3.20: \(3.20 = \frac{(2.0-2x)^2}{(0.25+x)(0.25-x)}\) This is a difficult equation to solve analytically, so we will make the assumption that x is small. This means that (0.25 + x) and (2.0 - 2x) are approximately 0.25 and 2.0, respectively. Now, we can solve for x: \(3.20 \approx \frac{(2.0)^2}{(0.25)(0.25-x)}\) \(x \approx 0.0212\)
06

Calculate New Equilibrium Concentrations

Now that we've found x, we can plug it back into the final row of the ICE table to find the new equilibrium concentrations. \[H_2] = 0.25 + x = 0.25 + 0.0212 = 0.2712\, M \[F_2] = 0.25 - x = 0.25 - 0.0212 = 0.2288\, M \[HF] = 2.0 - 2x = 2.0 - 2(0.0212) = 1.9576\, M The new equilibrium concentrations are approximately:\([H_2] = 0.2712 M\), \([F_2] = 0.2288 M\), and \([HF] = 1.9576 M\).

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Most popular questions from this chapter

In a study of the reaction $$ 3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g) $$ at \(1200 \mathrm{~K}\) it was observed that when the equilibrium partial pressure of water vapor is \(15.0\) torr, the total pressure at equilibrium is \(36.3\) torr. Calculate the value of \(K_{\mathrm{p}}\) for this reaction at \(1200 \mathrm{~K}\). (Hint: Apply Dalton's law of partial pressures.)

At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ At a low temperature, dry ice (solid \(\mathrm{CO}_{2}\) ), calcium oxide, and calcium carbonate are introduced into a \(50.0-\mathrm{L}\) reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C}\), resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2}\). For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900^{\circ} \mathrm{C} ?\) a. \(655 \mathrm{~g} \mathrm{CaCO}_{3}, 95.0 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=2.55 \mathrm{~atm}\) b. \(780 \mathrm{~g} \mathrm{CaCO}_{3}, 1.00 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\) c. \(0.14 \mathrm{~g} \mathrm{CaCO}_{3}, 5000 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\) d. \(715 \mathrm{~g} \mathrm{CaCO}_{3}, 813 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{~atm}\)

At a particular temperature, \(K=3.75\) for the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ If all four gases had initial concentrations of \(0.800 M\), calculate the equilibrium concentrations of the gases.

At a particular temperature, \(12.0\) moles of \(\mathrm{SO}_{3}\) is placed into a 3.0-L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ At equilibrium, \(3.0\) moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .{ }^{\circ} \mathrm{C}\). $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ If a \(20.0-\mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\) is put into a \(10.0\) - \(\mathrm{L}\) container and heated to \(800 .{ }^{\circ} \mathrm{C}\), what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?
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