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

Short Answer

Expert verified
\(K_{c}\) is the equilibrium constant with respect to concentration while \(K_{P}\) is the equilibrium constant in terms of partial pressure.

Step by step solution

01

Understanding \(K_{c}\)

The symbol \(K_{c}\) stands for the equilibrium constant with respect to concentration. It is a measure of the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients, at equilibrium.
02

Understanding \(K_{P}\)

\(K_{P}\) also stands for the equilibrium constant, but in terms of partial pressure. It is used when dealing with gases and is a measure of the ratio of the partial pressures of the products to the reactants, each raised to the power of their stoichiometric coefficients, at equilibrium.
03

Distinguishing between \(K_{c}\) and \(K_{P}\)

Both \(K_{c}\) and \(K_{P}\) are measures of the position of equilibrium for a chemical reaction. While the former uses concentrations (usually in mol/L), the latter uses partial pressures (usually in atmospheres), making it more appropriate for gases.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry that describes a state in which both the forward and backward reactions occur at the same rate. This means that the concentrations of the reactants and products remain constant over time, even though the reaction has not necessarily stopped.

Equilibrium is dynamic; it's not about the stoppage of reactions, but rather about achieving a balance in which the reactants and products form at an equal rate.

When discussing chemical equilibrium, we often refer to the equilibrium constant. This constant, denoted as either \(K_c\) or \(K_P\), helps chemists understand the proportion of reactants to products at equilibrium. This is crucial for predicting how changes in conditions such as concentration, pressure, or temperature will affect the system.
  • If \(K > 1\), the products are favored at equilibrium.
  • If \(K < 1\), the reactants are favored at equilibrium.
  • If \(K = 1\), neither reactants nor products are favored, indicating a perfect balance.
Reaction Quotient
The reaction quotient, symbolized as \(Q\), is a tool used to determine the direction in which a reaction will proceed in order to reach equilibrium. Like \(K_c\), it is calculated by taking the ratio of the concentrations of products to reactants.

The difference between \(Q\) and \(K\) is that \(Q\) applies to the reaction at any point in time, while \(K\) applies specifically at equilibrium. Comparing \(Q\) and \(K\) helps predict the direction of the reaction. If:
  • \(Q < K\): The reaction moves forward to produce more products.
  • \(Q > K\): The reaction shifts backward to produce more reactants.
  • \(Q = K\): The system is at equilibrium, and no net change occurs.
Thus, the reaction quotient serves as a valuable snapshot tool, helping chemists assess current conditions and predict how the system needs to adjust to achieve equilibrium.
Partial Pressure
Partial pressure is especially relevant in reactions involving gases. In a gaseous mixture, each gas contributes to the total pressure exerted by the mixture, which is referred to as its partial pressure.

Partial pressure is important for calculating the equilibrium constant in gaseous systems, denoted as \(K_P\). It enables the conversion of concentration-based equilibrium constants (\(K_c\)) to pressure-based ones using the ideal gas law: \[ K_P = K_c (RT)^{\Delta n} \] Here, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas (products minus reactants).

Understanding partial pressures allows scientists to predict how changes to pressure or volume can impact the equilibrium position, especially when dealing with gas-phase reactions. For example, increasing the pressure by compressing the gas mixture can shift the equilibrium position towards the side of the reaction with fewer moles of gas, as per Le Chatelier's principle. This adjustment helps maintain system balance.

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Most popular questions from this chapter

Consider the reaction $$ \begin{aligned} 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) & \\ \Delta H^{\circ}=&-198.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to \((\mathrm{a})\) increase the temperature, (b) increase the pressure, (c) increase \(\mathrm{SO}_{2},\) (d) add a catalyst, (e) add helium at constant volume.

Consider the heterogeneous equilibrium process: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ At \(700^{\circ} \mathrm{C}\), the total pressure of the system is found to be 4.50 atm. If the equilibrium constant \(K_{P}\) is 1.52 , calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\)

A sample of pure \(\mathrm{NO}_{2}\) gas heated to \(1000 \mathrm{~K}\) 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.

The decomposition of ammonium hydrogen sulfide $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ is an endothermic process. A \(6.1589-\mathrm{g}\) sample of the solid is placed in an evacuated 4.000 - \(L\) vessel at exactly \(24^{\circ} \mathrm{C}\). After equilibrium has been established, the total pressure inside is \(0.709 \mathrm{~atm}\). Some solid \(\mathrm{NH}_{4} \mathrm{HS}\) remains in the vessel. (a) What is the \(K_{P}\) for the reaction? (b) What percentage of the solid has decomposed? (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel?

Heating solid sodium bicarbonate in a closed vessel established this equilibrium: \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system, (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system, (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.

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