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Chapter 13: Problem 57

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.

Short Answer

Expert verified
The equilibrium concentrations of the gases are approximately: \([\mathrm{SO}_{2}]_\text{eqm} \approx 0.270\,\text{M}, [\mathrm{NO}_{2}]_\text{eqm} \approx 0.270\,\text{M}, [\mathrm{SO}_{3}]_\text{eqm} \approx 1.330\,\text{M}, [\mathrm{NO}]_\text{eqm} \approx 1.330\,\text{M}\).

Step by step solution

01

Write the equilibrium constant expression

The equilibrium constant expression (K) can be written for the given reaction as follows: \[K =\frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{\left[\mathrm{SO}_{2}\right][\mathrm{NO}_{2}]}\]
02

Set up an ICE (initial-change-equilibrium) table

Set up an ICE table to represent the initial concentrations, change in concentrations, and equilibrium concentrations of reactants and products in the reaction. | | SO₂(g) | NO₂(g) | ⇌ | SO₃(g) | NO(g) | |--------|--------|--------|---|--------|-------| |Initial | 0.800 | 0.800 | | 0.800 | 0.800 | | Change | -x | -x | | +x | +x | |Eqmbrium| 0.800-x| 0.800-x| | 0.800+x | 0.800+x| Here, x represents the change in molar concentrations of reactants and products during the reaction until equilibrium is reached.
03

Plug the equilibrium concentrations into the K expression

Substitute the equilibrium concentrations from the ICE table into the equilibrium constant expression. Here, K = 3.75. \[3.75 = \frac{(0.800+x)(0.800+x)}{(0.800-x)(0.800-x)}\]
04

Solve for x

Solve the resulting equation for x by clearing the fractions and simplifying. First, multiply both sides by \((0.800 - x)^2\) \[3.75(0.800-x)^2 = (0.800+x)^2\] Expand both sides of the equation: \[2.25 - 6x +3.75x^2 = 0.64 + 1.6x + x^2\] Now, rearrange this equation to form a quadratic equation: \[2.15x^2 - 7.6x +1.61 = 0\] Solve the quadratic equation for x using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\] Where \(a = 2.15, b = -7.6,\) and \(c = 1.61\) We get two possible values for x: \(x_1 \approx 0.530\) and \(x_2 \approx 1.40\). However, since the maximum possible value for x is 0.800 (initial concentration of reactants), we discard \(x_2\). Therefore, \(x \approx 0.530\).
05

Calculate the equilibrium concentrations

Plug the value of x back into the equilibrium expressions from the ICE table: \[ [\mathrm{SO}_{2}]_\text{eqm} = 0.800-x = 0.800 - 0.530 \approx 0.270 \text{M}\] \[ [\mathrm{NO}_{2}]_\text{eqm} = 0.800-x = 0.800 - 0.530 \approx 0.270 \text{M}\] \[ [\mathrm{SO}_{3}]_\text{eqm} = 0.800+x = 0.800 + 0.530 \approx 1.330 \text{M}\] \[ [\mathrm{NO}]_\text{eqm} = 0.800+x = 0.800 + 0.530 \approx 1.330 \text{M}\] So, the equilibrium concentrations of the gases are approximately: \([\mathrm{SO}_{2}] \approx 0.270\,\text{M}, [\mathrm{NO}_{2}] \approx 0.270\,\text{M}, [\mathrm{SO}_{3}] \approx 1.330\,\text{M}, [\mathrm{NO}] \approx 1.330\,\text{M}\).

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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.

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned} 2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\ P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\ P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\ P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm} \end{aligned}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

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

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.

At a particular temperature a \(2.00-\mathrm{L}\) flask at equilibrium contains \(2.80 \times 10^{-4}\) mole of \(\mathrm{N}_{2}, 2.50 \times 10^{-5}\) mole of \(\mathrm{O}_{2},\) and \(2.00 \times 10^{-2}\) mole of $\mathrm{N}_{2} \mathrm{O}\( . Calculate \)K$ at this temperature for the reaction $$2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{N}_{2} \mathrm{O}(g)$$ If $\left[\mathrm{N}_{2}\right]=2.00 \times 10^{-4} M,\left[\mathrm{N}_{2} \mathrm{O}\right]=0.200 M,\( and \)\left[\mathrm{O}_{2}\right]=\( \)0.00245 M,$ does this represent a system at equilibrium?
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