For the dissociation reaction \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{g})+\) \(\mathrm{S}_{2}(\mathrm{g}), K_{\mathrm{p}}=1.2 \times 10^{-2}\) at \(1065^{\circ} \mathrm{C} .\) For this same reaction at \(1000 \mathrm{K},\) (a) \(K_{\mathrm{c}}\) is less than \(K_{\mathrm{p}} ;\) (b) \(K_{\mathrm{c}}\) is greater than \(K_{\mathrm{p}} ;(\mathrm{c}) K_{\mathrm{c}}=K_{\mathrm{p}} ;\) (d) whether \(K_{\mathrm{c}}\) is less than, equal to, or greater than \(K_{\mathrm{p}}\) depends on the total gas pressure.

Short Answer

Expert verified
(b) \(K_c\) is less than \(K_p\).

Step by step solution

01

Identify the Reaction

The first step is to clearly identify the reaction. It is given as: \(2H_{2}S(g) \leftrightharpoons 2H_{2}(g) + S_{2}(g)\)
02

Calculate the Change in Moles

Determine the change in the number of moles \(\Delta n\). This is calculated by subtracting the total number of moles of gaseous reactants from the total number of moles of gaseous products. For this reaction, we have: \(\Delta n = (2 + 1) - 2 = 1\)
03

Apply the Relationship

Use the relationship between \(K_c\) and \(K_p\), which is \(K_p = K_c(RT)^{\Delta n}\), when \(\Delta n = 1\) and \(RT\) is always positive and greater than 1 at temperatures above absolute zero, \(K_p > K_c\). So the correct answer is (b) \(K_c\) is less than \(K_p\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant is a critical concept in chemistry that quantifies the ratio of concentrations of products to reactants at the point of chemical equilibrium. It is denoted as \(K\) and can take the form of \(K_c\) for concentrations or \(K_p\) for partial pressures when dealing with gases. Understanding \(K\) helps predict the extent of a reaction under given conditions. It's important to remember that the value of the equilibrium constant is dependent on temperature and will change if the temperature of the system is altered.

When a reaction has reached equilibrium, it doesn't mean the reactants and products are present in equal amounts, rather that their rates of formation are equal, leading to constant concentrations. The higher the value of \(K\), the more the equilibrium lies towards the products, indicating a 'product-favored' reaction. Conversely, a low \(K\) signals a 'reactant-favored' reaction where the formation of products is less favored.
Kp and Kc Relationship
The relationship between \(K_p\) and \(K_c\) is fundamental for dealing with gas-phase reactions. These equilibrium constants are related through the expression \(K_p = K_c(RT)^{\Delta n}\), where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas between products and reactants. It's important to highlight that \(K_p\) pertains to partial pressures and is used when the reaction involves gases, while \(K_c\) is based on molar concentrations.

In any chemical reaction where \(\Delta n\) is not zero, \(K_p\) will not be equal to \(K_c\). For instance, if \(\Delta n\) is positive, meaning there are more gaseous products than reactants, \(K_p\) will be greater than \(K_c\) if the temperatures are above absolute zero, due to the \(RT\) term being raised to a positive power. This relationship allows us to convert between the two constants provided we know the temperature and the change in gaseous moles.
Gas-Phase Reactions
Gas-phase reactions are chemical processes that occur between substances in the gaseous state. The behavior and study of these reactions are distinctly influenced by the properties of gases, such as volume, pressure, temperature, and concentration. \(K_p\) becomes particularly useful in these cases as it deals with the equilibrium conditions involving the partial pressures of the gases involved.

For gas-phase reactions, partial pressures serve as a proxy for concentration, since according to the ideal gas law \(PV = nRT\), pressure is directly proportional to the concentration. Moreover, the unique nature of gases allows us to use the ideal gas law to relate these partial pressures to molar concentrations and thus to convert between different expressions of the equilibrium constant. The understanding and calculation of reaction directions and extents in gas-phase reactions is not just academic; it's vital for practical applications such as industrial synthesis, environmental monitoring, and even in astrophysical chemistry.
Le Chatelier's Principle
Le Chatelier's principle provides a predictive view on how a system at equilibrium reacts to external changes. It states that if an external stress, such as changes in pressure, temperature, or concentration, is applied to a system in equilibrium, the system will adjust itself to counteract the imposition of the stress and restore a new equilibrium state. This principle enables chemists to control chemical reactions to some degree and optimize yields of desired products.

Application of Le Chatelier's principle is manifold. For example, increasing the concentration of reactants will typically shift the equilibrium towards the formation of more products. Conversely, a rise in temperature for an exothermic reaction will shift the balance towards the reactants, as heat is a product of the reaction and the system aims to absorb the excess heat by forming more reactants. Understanding this principle is essential for manipulating reaction conditions to achieve optimal results in both laboratory and industrial settings.

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Most popular questions from this chapter

At \(2000 \mathrm{K}, K_{c}=0.154\) for the reaction \(2 \mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If a \(1.00 \mathrm{L}\) equilibrium mixture at \(2000 \mathrm{K}\) contains \(0.10 \mathrm{mol}\) each of \(\mathrm{CH}_{4}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (a) what is the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) present? (b) Is the conversion of \(\mathrm{CH}_{4}(\mathrm{g})\) to \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) favored at high or low pressures? (c) If the equilibrium mixture at \(2000 \mathrm{K}\) is transferred from a 1.00 L flask to a 2.00 L flask, will the number of moles of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) increase, decrease, or remain unchanged?

Write the equilibrium constant expression for the following reaction, $$\begin{array}{r} \mathrm{Fe}(\mathrm{OH})_{3}+3 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ K=9.1 \times 10^{3} \end{array}$$ and compute the equilibrium concentration for \(\left[\mathrm{Fe}^{3+}\right]\) at \(\left.\mathrm{pH}=7 \text { (i.e., }\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7}\right)\)

Equilibrium is established in a 2.50 L flask at \(250^{\circ} \mathrm{C}\) for the reaction $$\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=3.8 \times 10^{-2}$$ How many moles of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present at equilibrium, if (a) 0.550 mol each of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are initially introduced into the flask? (b) \(0.610 \mathrm{mol} \mathrm{PCl}_{5}\) alone is introduced into the flask?

Is a mixture of \(0.0205 \mathrm{mol} \mathrm{NO}_{2}(\mathrm{g})\) and \(0.750 \mathrm{mol}\) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) in a \(5.25 \mathrm{L}\) flask at \(25^{\circ} \mathrm{C},\) at equilibrium? If not, in which direction will the reaction proceed toward products or reactants? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3} \mathrm{at} 25^{\circ} \mathrm{C}$$

A mixture consisting of \(0.150 \mathrm{mol} \mathrm{H}_{2}\) and \(0.150 \mathrm{mol} \mathrm{I}_{2}\) is brought to equilibrium at \(445^{\circ} \mathrm{C},\) in a 3.25 L flask. What are the equilibrium amounts of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and HI? $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\ \mathrm\ {at}\ 445^{\circ} \mathrm{C}$$

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