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Chapter 13: Problem 19

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

Short Answer

Expert verified
To differentiate between K, Kp, and Q: K is the equilibrium constant expressed in terms of concentrations (\( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)), while Kp is the equilibrium constant expressed in terms of partial pressures (\( K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \)). Both K and Kp describe the ratio of products to reactants at equilibrium. On the other hand, the reaction quotient, Q, can be calculated using both concentrations (\( Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)) or partial pressures (\( Q_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \)), but it represents the current state of the reaction, whether or not it is at equilibrium. Comparing Q to K or Kp helps determine the direction in which a reaction will proceed to reach equilibrium.

Step by step solution

01

Definition of K (Equilibrium Constant)

K, the equilibrium constant, is a dimensionless value that is used to describe the ratio of concentrations of products to reactants for a chemical reaction at equilibrium. It is defined by the ratio of the products' concentrations raised to their stoichiometric coefficients to the reactants' concentrations raised to their coefficients. For the general reaction: \( aA + bB \rightleftharpoons cC + dD \) The equilibrium constant (K) expression can be written as: \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \) Here, [A], [B], [C], and [D] represent the equilibrium concentrations of the respective species.
02

Definition of Kp (Equilibrium Constant in terms of Pressure)

Kp is the equilibrium constant expressed in terms of partial pressures instead of concentrations. For the same reaction we mentioned earlier: \( aA + bB \rightleftharpoons cC + dD \) The Kp expression can be written 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 species at equilibrium.
03

Definition of Q (Reaction Quotient)

Q, the reaction quotient, is similar to the equilibrium constant, but it is calculated using the current concentrations or partial pressures of the reactants and products, regardless of whether the system is at equilibrium or not. For the reaction we've been using: \( aA + bB \rightleftharpoons cC + dD \) The reaction quotient (Q) expression can be written as: \( Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \) (for concentrations) or \( Q_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \) (for partial pressures)
04

Distinguishing Between K, Kp, and Q

To summarize, K and Kp are both equilibrium constants, with K representing the equilibrium constant in terms of concentrations, and Kp representing the equilibrium constant in terms of partial pressures. The reaction quotient, Q, can be calculated in terms of concentrations or partial pressures as well but is used to represent the current state of a reaction, regardless of whether it has reached equilibrium. The value of Q can be compared to K or Kp to determine the direction in which a reaction will proceed to reach equilibrium.

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Most popular questions from this chapter

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. $2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)$ b. $2 \mathrm{NBr}_{3}(s) \Longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)$ c. $2 \mathrm{KClO}_{3}(s) \Longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$ d. $\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)$

The equilibrium constant \(K_{\mathrm{p}}\) is \(2.4 \times 10^{3}\) at a certain temperature for the reaction $$2 \mathrm{NO}(g) \leftrightharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those not at equilibrium, in which direction will the system shift? a. $P_{\mathrm{NO}}=0.012 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.11 \mathrm{atm}, P_{\mathrm{O}_{2}}=2.0 \mathrm{atm}$ b. $P_{\mathrm{NO}}=0.0078 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.36 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.67 \mathrm{atm}$ c. $P_{\mathrm{NO}}=0.0062 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.51 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.18 \mathrm{atm}$

A 1.00-L flask was filled with 2.00 moles of gaseous \(\mathrm{SO}_{2}\) and 2.00 moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K,\) for this reaction.

Consider the reaction $$\mathrm{P}_{4}(g) \rightleftharpoons 2 \mathrm{P}_{2}(g)$$ where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at 1325 \(\mathrm{K}\) . In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at 1325 \(\mathrm{K}\) , the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 -L vessel, calculate the equilibrium concentrations of all species.
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