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Chapter 13: Problem 63

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.0 moles of pure \(\mathrm{NOCl}\) in a 2.0 \(\mathrm{L}\) flask b. 1.0 mole of NOCl and 1.0 mole of NO in a 1.0 - flask c. 2.0 moles of \(\mathrm{NOCl}\) and 1.0 mole of \(\mathrm{Cl}_{2}\) in a \(1.0-\mathrm{L}\) flask

Short Answer

Expert verified
For scenario a, the equilibrium concentrations are \([\mathrm{NOCl}] \approx 0.974\, \mathrm{M}\), \([\mathrm{NO}] \approx 0.026\, \mathrm{M}\), and \([\mathrm{Cl_2}] \approx 1.3 \times 10^{-2}\, \mathrm{M}\).

Step by step solution

01

Set up an ICE table

For this scenario, consider the initial concentrations as follows: \[\mathrm{[NOCl]_0 = \frac{2.0\, moles}{2.0\, L} = 1.0\, M}\] \[\mathrm{[NO]_0 = [Cl_2]_0 = 0}\] Create an ICE table: | | \(\mathrm{NOCl}\) | \(\mathrm{NO}\) | \(\mathrm{Cl_2}\) | |--------|---------|-------|---------| | Initial| 1.0 M | 0 | 0 | | Change | -2x | +2x | +x | | Equil. | 1.0-2x | 2x | x |
02

Write the equilibrium expression

Write the equation for the equilibrium constant: \[K=\frac{[\mathrm{NO}]^2[\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
03

Substitute values and solve for x

Substitute the values from the ICE table into the equilibrium expression: \[1.6 \times 10^{-5} = \frac{(2x)^2(x)}{(1.0-2x)^2}\] Solve the quadratic equation for x. Since the equilibrium constant is very small, we can assume that \(2x \ll 1\) and simplify to: \[ 1.6 \times 10^{-5} \approx \frac{4x^3}{1} \] Solving for x gives: \[ x \approx 1.3 \times 10^{-2} \]
04

Find equilibrium concentrations

Now calculate the equilibrium concentrations: \[ [\mathrm{NOCl}]_{eq} = 1.0 - 2x \approx 1.0 - 2(1.3 \times 10^{-2}) \approx 0.974\, \mathrm{M} \] \[ [\mathrm{NO}]_{eq} = 2x \approx 2(1.3 \times 10^{-2}) \approx 0.026\, \mathrm{M} \] \[ [\mathrm{Cl_2}]_{eq} = x \approx 1.3 \times 10^{-2}\, \mathrm{M} \] So, for scenario a, the equilibrium concentrations are approximately \([\mathrm{NOCl}] \approx 0.974\, \mathrm{M}\), \([\mathrm{NO}] \approx 0.026\, \mathrm{M}\), and \([\mathrm{Cl_2}] \approx 1.3 \times 10^{-2}\, \mathrm{M}\). Repeat the same process for scenarios b and c.

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Most popular questions from this chapter

For the reaction $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),$ consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or \((b)\) you \(\operatorname{mix} 1.5\) moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$ b. $\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$ c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$

In which direction will the position of the equilibrium $$2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)$$ be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\operatorname{HI}(g)\) is removed. d. In a rigid reaction container, some Ar(g) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).

Consider the reaction $$\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q)$$ How will the equilibrium position shift if a. water is added, doubling the volume? b. \(\operatorname{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? [Fe(OH) \(_{3}\) is insoluble. \(]\) d. Fe(NO \(_{3} )_{3}(a q)\) is added?

The partial pressures of an equilibrium mixture of $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( and \)\mathrm{NO}_{2}(g)\( are \)P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.34\( atm and \)P_{\mathrm{NO}_{2}}=1.20 \mathrm{atm}$ at a certain temperature. The volume of the container is doubled. Calculate the partial pressures of the two gases when a new equilibrium is established.
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