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Chapter 14: Problem 94

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)?

Short Answer

Expert verified
(a) \(K_p\) will be a value calculated using the formula provided, after substitifying the known values. (b) Value of \(K_c\) for the reverse reaction is the reciprocal of the original \(1/1.2\), so it would be approximately 0.833. (c) Value of \(K_c\) for the halved reaction is the square root of the original \(1.2^{1/2}\), or approximately 1.095. (d) The values of \(K_p\) for the reactions described in (b) and (c) can be found similarly to in part (a).

Step by step solution

01

Part (a): Calculate \(K_p\)

We know that the relationship between \(K_c\) and \(K_p\) is given by \[K_p=K_c(RT)^{\Delta n}\] where \(\Delta n\) is change in moles of gaseous reactants and products (= moles of gaseous products - moles of gaseous reactants). Here, \(\Delta n = (2) - (1+3) = -2\). Substitute known values into the formula to compute \(K_p\).
02

Part (b): Find new \(K_p\)

When reversing a reaction, the equilibrium constant gets inverted. So, \(K_c\) for \(2\mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) is simply \(1/1.2\).
03

Part (c): Compute \(K_c\) for the given reaction

The given reaction is half of the original one (\(\frac{1}{2}\mathrm{N}_{2}(g)+\frac{3}{2}\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\)). When a reaction is multiplied by a factor (1/2 in this case), the equilibrium constant is raised to that factor. So, the new \(K_c\) is \(1.2^{1/2}\).
04

Part (d): Find \(K_p\) for both reactions in (b) & (c)

The values of \(K_p\) for the reactions in (b) and (c) can be found similarly to Step 1, realizing that for reaction (b), the value of \(\Delta n\) will reverse sign and for reaction (c), \(\Delta n\) remains the same given that the reaction is scaled, but not changed.

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Most popular questions from this chapter

Consider the following equilibrium process at \(700^{\circ} \mathrm{C}:\) $$2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g)$$ Analysis shows that there are 2.50 moles of \(\mathrm{H}_{2}\), \(1.35 \times 10^{-5}\) mole of \(\mathrm{S}_{2}\), and 8.70 moles of \(\mathrm{H}_{2} \mathrm{~S}\) present in a 12.0-L flask. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

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 \(\left(K_{\mathrm{c}}\right)\) for the reaction $$2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)$$ is \(4.17 \times 10^{-34}\) at \(25^{\circ} \mathrm{C} .\) What is the equilibrium constant for the reaction $$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$$ at the same temperature?

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 \(K_{P}\).

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.
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