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Chapter 14: Problem 42

At \(1000 \mathrm{~K},\) a sample of pure \(\mathrm{NO}_{2}\) gas 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
The pressure of \(NO_2\) at equilibrium is \(3 - 0.25 = 2.75\) atm, and the pressure of \(NO\) at equilibrium is \(0.5\) atm.

Step by step solution

01

Initial guess of partial pressures

Let's say that initially, before any decomposition, the pressure of \(NO_2\) is \(P\). As the reaction progresses to reach equilibrium, \(NO_2\) decomposes to form \(NO\) and \(O_2\). We know that 2 moles of \(NO_2\) give 2 moles of \(NO\) and 1 mole of \(O_2\). Let's consider that \(x\) is the amount of \(NO_2\) that dissociates at equilibrium. So, the change in pressure for \(NO_2\) is \(-x\), for \(NO\) it's \(+2x\) and for \(O_2\) it's \(+x\). Thus, at equilibrium, the pressures are: \(NO_2 (P-x)\), \(NO(2x)\), and \(O_2(x)\). But we know that the partial pressure of \(O_2\) at equilibrium is 0.25 atm, so we can say that \(x = 0.25\).
02

Calculate the pressure of \(NO_2\) at equilibrium

Substitute \(x = 0.25\) into the calculated expression for \(NO_2\) equilibrium pressure. So, \(NO_2\) pressure at equilibrium is \(P - 0.25\).
03

Apply the definition of the equilibrium constant \(K_P\)

The expression for the equilibrium constant \(K_P\) of the reaction is \[ K_P = \frac{{(P_{NO})^2 \cdot P_{O_2}}}{{(P_{NO_2})^2}} \] At equilibrium, we replace each \(P\) with the corresponding calculated equilibrium pressures. Therefore, \[ 158 = \frac{{(2 \cdot 0.25)^2 \cdot 0.25}}{{(P - 0.25)^2}} \]
04

Solve for \(P\)

Solve the equation above for \(P\), which gives the initial pressure of \(NO_2\). After solving the equation, we find that \(P = 3\) atm.
05

Calculate the pressure of \(NO\) at equilibrium

Substitute \(x = 0.25\) into the calculated expression for \(NO\) equilibrium pressure. So, \(NO\) pressure at equilibrium is \(2 \cdot 0.25 = 0.5\) atm.

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Most popular questions from this chapter

For the reaction $$\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g)$$ at \(700^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.534 .\) Calculate the number of moles of \(\mathrm{H}_{2}\) that are present at equilibrium if a mixture of 0.300 mole of \(\mathrm{CO}\) and \(0.300 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}\) is heated to \(700^{\circ} \mathrm{C}\) in a 10.0 -L container.

The equilibrium constant \(K_{\mathrm{c}}\) for the following reaction is 1.2 at \(375^{\circ} \mathrm{C}\). $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ (a) What is the value of \(K_{P}\) for this reaction? (b) What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?\) (c) What is the value of \(K_{c}\) for \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) \(\rightleftharpoons \mathrm{NH}_{3}(g) ?\) (d) What are the values of \(K_{P}\) for the reactions described in (b) and (c)?

Consider the following reaction at equilibrium: $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ From the data shown here, calculate the equilibrium constant (both \(K_{P}\) and \(K_{\mathrm{c}}\) ) at each temperature. Is the reaction endothermic or exothermic? $$ \begin{array}{clr} \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & {[\mathrm{A}](M)} & {[\mathrm{B}](M)} \\ 200 & 0.0125 & 0.843 \\ 300 & 0.171 & 0.764 \\ 400 & 0.250 & 0.724 \end{array} $$

A quantity of 0.20 mole of carbon dioxide was heated to a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ Under these conditions, the average molar mass of the gases was \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is \(K_{P}\) if the total pressure is 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.

The following equilibrium constants were determined at \(1123 \mathrm{~K}\) $$\begin{array}{ll}\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) & K_{P}^{\prime}=1.3 \times 10^{14} \\\\\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) & K_{P}^{\prime \prime}=6.0 \times 10^{-3}\end{array}$$ Write the equilibrium constant expression \(K_{P}\), and calculate the equilibrium constant at \(1123 \mathrm{~K}\) for $$\mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$
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