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Chapter 13: Problem 51

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 for the given reaction are: \(\mathrm{SO}_{2}\): \(0.545M\), \(\mathrm{NO}_{2}\): \(0.545M\), \(\mathrm{SO}_{3}\): \(1.055M\), and \(\mathrm{NO}\): \(1.055M\).

Step by step solution

01

Write the equilibrium expression

To start, we need to write an equilibrium expression for the given chemical reaction. The equilibrium expression is the ratio of the product concentrations to the reactant concentrations. Each concentration is raised to the power of its stoichiometric coefficient. In this case, the equilibrium expression is: $$ K = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]} $$
02

Set up the ICE table

An ICE table (Initial, Change, Equilibrium) helps to organize the information in a systematic way. For this problem, the ICE table would look like: | | Initial | Change | Equilibrium | |:-:|---------|--------|-------------| | SO2 | 0.800 | -x | 0.800-x | | NO2 | 0.800 | -x | 0.800-x | | SO3 | 0.800 | +x | 0.800+x | | NO | 0.800 | +x | 0.800+x | The change in the concentrations is represented by "x". The reactants' concentrations decrease by x, while the products' concentrations increase by x.
03

Substitute equilibrium concentrations into the equilibrium expression

We now substitute the equilibrium concentrations into the equilibrium expression we derived in Step 1: $$ 3.75 = \frac{(0.800+x)(0.800+x)}{(0.800-x)(0.800-x)} $$
04

Solve for x

Next, we solve for x using quadratic equation or other algebraic methods. For the given equilibrium constant, the equation simplifies as: $$ 3.75 = \frac{(0.800+x)^{2}}{(0.800-x)^{2}} $$ Now, take the square root of both sides: $$ \sqrt{3.75} = \frac{0.800+x}{0.800-x} $$ $$ 1.935 = \frac{0.800+x}{0.800-x} $$ We can now cross multiply and solve for x: $$ 1.935(0.800-x)=(0.800+x) $$ $$ 1.548 - 1.935x = 0.800+x $$ $$ 2.935x = 0.748 $$ $$ x = 0.255 $$
05

Calculate the equilibrium concentrations

Finally, we can plug in x to calculate the equilibrium concentrations: $$ [\mathrm{SO}_{2}] = 0.800 - 0.255 = 0.545 M $$ $$ [\mathrm{NO}_{2}] = 0.800 - 0.255 = 0.545 M $$ $$ [\mathrm{SO}_{3}] = 0.800 + 0.255 = 1.055 M $$ $$ [\mathrm{NO}] = 0.800 + 0.255 = 1.055 M $$ Now, we have the equilibrium concentrations for all gases: SO2: 0.545 M NO2: 0.545 M SO3: 1.055 M NO: 1.055 M

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Most popular questions from this chapter

Given the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\), consider the following situations: i. You have \(1.3 M \mathrm{~A}\) and \(0.8 \mathrm{M} \mathrm{B}\) initially. ii. You have \(1.3 M \mathrm{~A}, 0.8 M \mathrm{~B}\), and \(0.2 M \mathrm{C}\) initially. iii. You have \(2.0 \mathrm{M} \mathrm{A}\) and \(0.8 \mathrm{M} \mathrm{B}\) initially. Order the preceding situations in terms of increasing equilibrium concentration of D. Explain your order. Then give the order in terms of increasing equilibrium concentration of \(\mathrm{B}\) and explain.

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NBr}_{3}(s) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\) c. \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) d. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)

A \(1.604-\mathrm{g}\) sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and \(6.400 \mathrm{~g}\) oxygen gas are sealed into a 2.50-L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is \(0.326 \mathrm{~atm}\), and the pressure of water vapor is \(4.45\) atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

Nitric oxide and bromine at initial partial pressures of \(98.4\) and \(41.3\) torr, respectively, were allowed to react at \(300 . \mathrm{K}\). At equilibrium the total pressure was \(110.5\) torr. The reaction is $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), both at an initial partial pressure of \(0.30 \mathrm{~atm}\), were allowed to come to equilibrium at this temperature?

Explain the difference between \(K, K_{\mathrm{p}}\), and \(Q\).
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