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Chapter 14: Problem 52

How is an equilibrium constant different from a reaction quotient?

Short Answer

Expert verified
Answer: The main difference between an equilibrium constant (K) and a reaction quotient (Q) is that K represents the ratio of the concentrations or partial pressures of the products to the reactants in a chemical reaction at equilibrium, while Q represents the ratio at any point in time during the reaction before equilibrium is reached. By comparing the values of Q and K, we can determine the direction in which a reaction will proceed: if Q > K, the reaction will proceed in the reverse direction; if Q < K, the reaction will proceed in the forward direction; and if Q = K, the reaction is at equilibrium.

Step by step solution

01

Definition of Equilibrium Constant

An equilibrium constant (K) is a ratio of the concentrations or partial pressures of the products to the reactants in a chemical reaction at equilibrium. It is a measure of how far a reaction proceeds and is specific to a particular reaction at a given temperature. The equilibrium constant indicates the extent to which a reaction will reach completion. For a general reaction: Aa + Bb 鈫� Cc + Dd The equilibrium constant expression is given by: K = ([C]^c [D]^d) / ([A]^a [B]^b)
02

Definition of Reaction Quotient

A reaction quotient (Q) is a ratio similar to the equilibrium constant but is calculated using the initial concentrations or partial pressures of the reactants and products at any point in time before the system has reached equilibrium. It is a tool that allows us to monitor the progress of a reaction and predict the direction in which the reaction will proceed to achieve equilibrium. For the same general reaction as above, the reaction quotient expression is also given by: Q = ([C]^c [D]^d) / ([A]^a [B]^b)
03

Comparing K and Q

Both K and Q have the same form, where they are the ratios of concentrations or partial pressures of products to reactants. However, their main difference lies in when they are calculated. While K represents this ratio at equilibrium, Q represents this ratio at any point in time during the reaction. To determine the direction in which a reaction will proceed, we compare the values of Q and K: - If Q > K, the reaction will proceed in the reverse direction, converting products into reactants. - If Q < K, the reaction will proceed in the forward direction, transforming reactants into products. - If Q = K, the reaction is at equilibrium, meaning the rates of the forward and reverse reactions are equal.
04

Example

Consider the reaction: N2(g) + 3H2(g) 鈫� 2NH3(g) The equilibrium constant, K, for this reaction at a specific temperature is 1.45 x 10^5. At a given time, the concentrations of N2, H2, and NH3 are found to be 0.2 M, 0.01 M, and 2.0 M, respectively. To compare Q and K, we first calculate Q: Q = ([NH3]^2) / ([N2][H2]^3) = (2.0^2) / (0.2 x 0.01^3) 鈮� 8 x 10^6 In this case, Q > K, so the reaction will proceed in the reverse direction, converting NH3 back into N2 and H2 until equilibrium is reached.

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Most popular questions from this chapter

Could the quadratic equation be used to solve for the equilibrium concentration of \(\mathrm{NO}_{2}\) in the following reaction? $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$

Carbon Monoxide Poisoning Patients suffering from carbon monoxide poisoning are treated with pure oxygen to remove CO from the hemoglobin (Hb) in their blood. The two relevant equilibria are $$ \begin{aligned} \mathrm{Hb}+4 \mathrm{CO}(g) & \rightleftharpoons \mathrm{Hb}(\mathrm{CO}) \\ \mathrm{Hb}+4 \mathrm{O}_{2}(g) & \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4} \end{aligned} $$ The value of the equilibrium constant for CO binding to Hb is greater than that for \(\mathrm{O}_{2}\). How, then, does this treatment work?

Henry's law (Chapter 10 ) predicts that the solubility of a gas in a liquid increases with its partial pressure. Explain Henry's law in relation to Le Chatelicr's principle.

Hydrogen Production The steam-methane reforming reaction plays a key role in producing hydrogen gas for use as a fuel and as a reactant in ammonia production. The equilibrium constant \(\left(K_{\mathrm{p}}\right)\) of the reaction: \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 3 \mathrm{H}_{2}(g)+\mathrm{CO}(g)\) is 13.0 at \(700^{\circ} \mathrm{C}\) a. Describe two advantages in running this endothermic \(\left(\Delta H_{r x n}=206 \mathrm{kJ}\right)\) reaction at \(700^{\circ} \mathrm{C}\) instead of \(100^{\circ} \mathrm{C}\) b. If the initial partial pressures of the two reactants are \(\operatorname{cach} 5.00\) atm at \(700^{\circ} \mathrm{C}\) and no products are present, what is the partial pressurc of \(\mathrm{H}_{2}\) gas after cquilibrium is achicved?

At a temperature of \(1000 \mathrm{K}, \mathrm{SO}_{2}(g)\) combines with oxygen to make \(\mathrm{SO}_{3}(\mathrm{g})\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ Under these conditions, \(K_{p}=3.4\) a. Use the appropriate thermodynamic data in Appendix 4 to calculate the value of \(\Delta H_{\mathrm{ren}}^{\circ}\) for this reaction. b. What is the value of \(K_{p}\) for this reaction at \(298 \mathrm{K}\) ? c. Use the answer from part (b) to calculate the value of \(\Delta G_{\max }^{*}\) at \(298 \mathrm{K},\) and compare it to the value you obtaincd using the \(\Delta G_{\mathrm{f}}^{s}\) valucs in Appendix 4.
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