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Chapter 15: Problem 32

A sample of pure \(\mathrm{NO}_{2}\) gas heated to \(1000 \mathrm{~K}\) decomposes: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{P}\) is \(158 .\) Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is 0.25 atm at equilibrium. Calculate the pressure of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in the mixture.

Short Answer

Expert verified
\nThe pressures of \(NO_{2}\) and \(NO\) in the mixture are both approximately 0.02 atm.

Step by step solution

01

Understand the relationship through the reaction formula

The balanced chemical equation is: \(2NO_{2} \rightleftharpoons 2NO + O_{2}\). For each 2 moles of \(NO_{2}\) that decompose, 2 moles of \(NO\) and one mole of \(O_{2}\) are produced.
02

Set up the equilibrium expression

According to the reaction, the equilibrium constant expression for the reaction in terms of partial pressures (This is why it's called \(K_{P}\)) would be: \( K_{P} = \frac{P_{NO}^{2} \cdot P_{O_{2}}}{P_{NO_{2}}^{2}} \). Here, \(P_{NO}\), \(P_{O_{2}}\) and \(P_{NO_{2}}\) represent the partial pressures of \(NO\), \(O_{2}\) and \(NO_{2}\) respectively.
03

Express \(P_{NO}\) in terms of \(P_{NO_{2}}\)

Given that the partial pressure of \(O_{2}\) is 0.25 atm, you can plug this into the equilibrium expression. Furthermore, when equilibrium is reached, the pressure of \(NO_{2}\) that has decomposed will be equal to the pressure of \(NO\) that has been produced. This means we can express \(P_{NO}\) in terms of \(P_{NO_{2}}\). From the balanced chemical reaction, we get the same amount of \(NO_{2}\) and \(NO\), so we can say \(P_{NO} = P_{NO_{2}}\). After substitution, the equation becomes: \( K_{P} = \frac{P_{NO_{2}}^{2} \cdot P_{O_{2}}}{P_{NO_{2}}^{2}} = P_{O_{2}} \)
04

Solve the equation to get \(P_{NO_{2}}\) and \(P_{NO}\)

Now, we solve the equation for \(P_{NO_{2}}\), and given that \(K_P = 158\) and \(P_{O_{2}} = 0.25\) atm, we get: \(P_{NO_{2}} = \sqrt{\frac{P_{O_{2}}}{K_P}} = \sqrt{\frac{0.25}{158}}\). Both \(P_{NO_{2}}\) and \(P_{NO}\) have the same value.

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Most popular questions from this chapter

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$ \mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g) $$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

In the uncatalyzed reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ at \(100^{\circ} \mathrm{C}\) the pressures of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.377 \mathrm{~atm}\) and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm} .\) What would happen to these pressures if a catalyst were present?

Baking soda (sodium bicarbonate) undergoes thermal decomposition as $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

Explain the difference between physical equilibrium and chemical equilibrium. Give two examples of each.

What do the symbols \(K_{c}\) and \(K_{P}\) represent?
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