Explain the difference between \(K, K_{\mathrm{p}}\), and \(Q\).

Short Answer

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In short, \(K\) and \(K_{p}\) are both equilibrium constants for chemical reactions, with \(K\) defined in terms of concentrations and \(K_{p}\) in terms of partial pressures for gas-phase reactions. They express the extent to which a reaction proceeds before reaching equilibrium. \(Q\) is the reaction quotient, calculated at any point during a reaction, and is used to predict the direction in which the reaction will proceed to reach equilibrium. Comparing \(Q\) with \(K\) or \(K_{p}\) provides insights on whether the reaction needs to proceed toward products or reactants to achieve equilibrium.

Step by step solution

01

Definition of K (Equilibrium Constant)

K, or the equilibrium constant, is a number that expresses how far a chemical reaction proceeds to the right (toward products) before reaching a state of equilibrium. It is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. Mathematically, for a general balanced equation: \[aA + bB \rightleftharpoons cC + dD\] The equilibrium constant K is defined as: \[K = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] where [A], [B], [C], and [D] represent the equilibrium concentrations (in mol/L) of the respective species.
02

Definition of K_p (Equilibrium Constant in terms of Pressure)

For gaseous reactions, the equilibrium constant can also be expressed in terms of the partial pressures of the reactants and products involved in the reaction. The equilibrium constant K_p is defined as the equilibrium constant in terms of pressure for gas-phase reactions. Mathematically, for the same balanced equation: \[aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\] K_p is defined as: \[K_{p} = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}\] Here, P_A, P_B, P_C, and P_D represent the partial pressures of the respective gaseous species (in atm) at equilibrium. The relationship between K and K_p is given by the equation: \[K_p = K(RT)^{\Delta n}\] Here, R is the universal gas constant, T is the temperature in Kelvin, and Δn represents the change in the number of moles of gas for the reaction (Δn = c + d - a - b).
03

Definition of Q (Reaction Quotient)

The reaction quotient, denoted by Q, is a number that gives a snapshot of a reaction's progress at any given moment. Like K and K_p, it is calculated using either concentrations or partial pressures of the reactants and products. However, Q is not specific to equilibrium conditions. Q can be calculated at any point during the reaction and is used to predict the direction in which the reaction will proceed to reach equilibrium. Mathematically, for the example reaction: \[aA + bB \rightleftharpoons cC + dD\] Q is defined as: \[Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] For any given moment during the reaction, comparing Q with K or K_p will provide insights on the direction of the reaction: - If Q < K (or K_p), it means that the reaction will proceed to the right (toward products) to achieve equilibrium. - If Q > K (or K_p), the reaction will proceed to the left (toward reactants) to achieve equilibrium. - If Q = K (or K_p), it indicates that the reaction is already at equilibrium. In summary, K and K_p are both equilibrium constants, with K defined in terms of concentrations and K_p defined in terms of partial pressures for gas-phase reactions. On the other hand, Q is the reaction quotient calculated at any point during a reaction to predict the direction of the reaction to reach equilibrium.

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Most popular questions from this chapter

The compound \(\mathrm{VCl}_{4}\) undergoes dimerization in solution: $$ 2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8} $$ When \(6.6834 \mathrm{~g} \mathrm{VCl}_{4}\) is dissolved in \(100.0 \mathrm{~g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\). Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(1.696\) \(\mathrm{g} / \mathrm{cm}^{3}\), and \(K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol}\) for \(\mathrm{CCl}_{4} .\).

For the reaction $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3} M\), \(\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{~N}_{2}(g)\right]=5.3 \times 10^{-2} M\), and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=\) \(2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

An \(8.00-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$ \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ At equilibrium the total pressure and the density of the gaseous mixture were \(1.80 \mathrm{~atm}\) and \(1.60 \mathrm{~g} / \mathrm{L}\), respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

Suppose the reaction system $$ \mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional \(\mathrm{UO}_{2}(s)\) is added to the system. b. The reaction is performed in a glass reaction vessel; \(\mathrm{HF}(g)\) attacks and reacts with glass. c. Water vapor is removed.

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ The initial density of the system was recorded as \(4.495 \mathrm{~g} / \mathrm{L}\). After equilibrium was reached, the density was noted to be \(4.086 \mathrm{~g} / \mathrm{L}\). a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K ?\) Explain each answer.

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