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Chapter 14: Problem 6

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

Short Answer

Expert verified
\(K_{\mathrm{c}}\) represents the equilibrium constant in terms of molar concentrations, used for aqueous solutions and gases. \(K_{P}\) represents the equilibrium constant in terms of partial pressures, applied for gas-phase reactions.

Step by step solution

01

Interpreting \(K_{\mathrm{c}}\)

\(K_{\mathrm{c}}\) is the equilibrium constant in terms of molar concentrations of the reactants and products. It is used when the substances in the chemical reaction are present in aqueous solution or in the gaseous state. It does not include pure solids or pure liquids because their concentrations are invariant.
02

Interpreting \(K_{P}\)

\(K_{P}\) is the equilibrium constant given in terms of partial pressures of gases involved in the equilibrium reaction. It is used when dealing with gas-phase reactions. The partial pressures are usually given in atmospheres.
03

Key Differences between \(K_{\mathrm{c}}\) and \(K_{P}\)

Though both \(K_{\mathrm{c}}\) and \(K_{P}\) represent chemical equilibrium constants, they are used in different contexts which mainly depends on the state of the reactants and products involved in the chemical reaction. \(K_{\mathrm{c}}\) is used when dealing with aqueous solutions and gases in terms of molar concentration, while \(K_{P}\) is used for gas-phase reactions in terms of partial pressures.

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Most popular questions from this chapter

Explain Le Châtelier's principle. How can this principle help us maximize the yields of reactions?

The equilibrium constant \(K_{P}\) for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures \(0.177 \mathrm{~atm},\) 0.223 atm, and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose \(\rightleftharpoons\) glucose A chemist prepared a \(0.244 M\) fructose solution at \(25^{\circ} \mathrm{C}\). At equilibrium, it was found that its concentration had decreased to \(0.113 M .\) (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose?

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=260 \mathrm{~kJ} /\mathrm{mol}\end{array}$$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\), in the presence of air, to convert the remaining methane to hydrogen: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

Consider the following reaction at \(1600^{\circ} \mathrm{C}\). $$\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)$$ When 1.05 moles of \(\mathrm{Br}_{2}\) are put in a 0.980 - \(\mathrm{L}\) flask, 1.20 percent of the \(\mathrm{Br}_{2}\) undergoes dissociation. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.
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