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Chapter 14: Problem 64

Pure nitrosyl chloride (NOCl) gas was heated to \(240^{\circ} \mathrm{C}\) in a \(1.00-\mathrm{L}\) container. At equilibrium the total pressure was 1.00 atm and the \(\mathrm{NOCl}\) pressure was 0.64 atm. $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate the partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) in the system. (b) Calculate the equilibrium constant \(K_{P}\).

Short Answer

Expert verified
The partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) at equilibrium are 0.36 atm and 0.18 atm respectively. The equilibrium constant \(K_{P}\) for the reaction at \(\mathrm{240^{\circ} C}\) is 0.15625.

Step by step solution

01

Analyze the reaction and given data

Firstly, write down the stoichiometrically balanced chemical reaction, as it is given: \(2 \mathrm{~NOCl}(g) \rightleftharpoons 2 \mathrm{~NO}(g)+\mathrm{Cl}_{2}(g)\). Analyze the given data. It is known that the total pressure at equilibrium is 1.00 atm, and the \(\mathrm{NOCl}\) pressure is 0.64 atm.
02

Calculate partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\)

In the equilibrium reaction, 2 moles of \(\mathrm{NOCl}\) decompose to form 2 moles of \(\mathrm{NO}\) and 1 mole of \(\mathrm{Cl}_{2}\). This implies that for every 1 atm of pressure drop in \(\mathrm{NOCl}\), there should be an increase of 0.5 atm in \(\mathrm{Cl}_{2}\) and 1 atm in \(\mathrm{NO}\), keeping the total pressure at equilibrium constant. Mathematically, \(P_{\mathrm{NO}} = P_{\mathrm{NOCl, initial}} - P_{\mathrm{NOCl, final}} = 1.00~\mathrm{atm} - 0.64~\mathrm{atm} = 0.36~\mathrm{atm}\) and \(P_{\mathrm{Cl}_{2}} = P_{\mathrm{NO}}/2 = 0.36~\mathrm{atm}/2 = 0.18~\mathrm{atm}\). So, the partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) are 0.36 atm and 0.18 atm respectively.
03

Calculate the equilibrium constant \(K_{P}\)

The equilibrium constant \(K_{P}\) is the ratio of the product of the partial pressures of the products to the product of the partial pressures of the reactants, each raised to a power equal to its stoichiometric coefficient. Based on the balanced reaction equation, we have: \(K_{P} = \frac{(P_{\mathrm{NO}})^{2} \cdot P_{\mathrm{Cl}_{2}}}{(P_{\mathrm{NOCl}})^{2}} = \frac{(0.36)^{2} \cdot 0.18}{(0.64)^{2}} = 0.15625\). The value of \(K_{P}\) is 0.15625.

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Most popular questions from this chapter

Consider the dissociation of iodine: $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ A 1.00-g sample of \(I_{2}\) is heated to \(1200^{\circ} \mathrm{C}\) in a \(500-\mathrm{mL}\) flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in \(14.117(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.]

The equilibrium constant \(\left(K_{P}\right)\) for the reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) is 2.93 at \(127^{\circ} \mathrm{C}\) Initially there were 2.00 moles of \(\mathrm{PCl}_{3}\) and 1.00 mole of \(\mathrm{Cl}_{2}\) present. Calculate the partial pressures of the gases at equilibrium if the total pressure is 2.00 atm.

At \(1130^{\circ} \mathrm{C}\) the equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)$$ is \(2.25 \times 10^{-4}\). If \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=4.84 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=\) \(1.50 \times 10^{-3} M,\) calculate \(\left[\mathrm{S}_{2}\right]\)

Consider the following reaction, which takes place in a single elementary step: $$2 \mathrm{~A}+\mathrm{B} \underset{k_{-1}}{\frac{k_{1}}{\longrightarrow}} \mathrm{A}_{2} \mathrm{~B}$$ If the equilibrium constant \(K_{\mathrm{c}}\) is 12.6 at a certain temperature and if \(k_{\mathrm{r}}=5.1 \times 10^{-2} \mathrm{~s}^{-1},\) calculate the value of \(k_{\mathrm{f}}\).

The "boat" form and "chair" form of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) interconverts as shown here: In this representation, the \(\mathrm{H}\) atoms are omitted and a \(\mathrm{C}\) atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each direction. The activation energy for the chair \(\longrightarrow\) boat conversion is \(41 \mathrm{~kJ} / \mathrm{mol} .\) If the frequency factor is \(1.0 \times 10^{12} \mathrm{~s}^{-1},\) what is \(k_{1}\) at \(298 \mathrm{~K} ?\) The equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(9.83 \times 10^{3}\) at \(298 \mathrm{~K}\).
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