Reaction Direction Comparing Q and

Suppose you start a reaction with a mixture of reactants and products and you know the equilibrium constant at the temperature of the reaction. Because the value of Q can change, depending on the initial concentrations, Q can be smaller than K, larger than K, or, when the system reaches equilibrium, equal to K. By comparing the value of Q at a particular time with the known K, you can tell whether the reaction has attained equilibrium or, if not, in which direction it is progressing. With product terms in the numerator of Q and reactant terms in the denominator, more product makes Q larger, and more reactant makes Q smaller. 

Reactants----- Products Equilibrium no net change Reactants -------Products 

Plan We write the expression for Q, find its value by substituting the given concentrations, and then compare its value with the given Kc. 

With Qc Kc, the reaction is not at equilibrium and will proceed to the left until Qc = Kc-Check With [NO2] [N2O4], we expect to obtain a value for Qc that is greater than 0.21. If Qc Kc, the numerator will decrease and the denominator will increase until Qc = Kc, that is, this reaction will proceed toward reactants. 

FOLLOW-UP PROBLEM 17.3 Chloromethane forms by the reaction CH4(g) + Cl2(g) CHjClCg) -I- HCl(g) 

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A Figure 15.8 Predicting the direction of a reaction by comparing Q and ff at a given temperature. 

Strategy First calculate the partial pressures of N204 and N02, using the ideal gas law as applied to mixtures P, = tiiRT/V. Then calculate Q. Finally, compare Q and K to predict the direction of reaction. 

Since reactions always move toward equilibrium, Q will always change toward K. Thus we can compare Q and K for a reaction at any given moment, and learn in which direction the reaction will proceed. 

At any point in a reaction, we can predict its direction by comparing Q and K if Q< K, the reaction will form more product if Q > K, the reaction will form more reactant if Q = K, the reaction is at equilibrium. (Section 17.4)... 

For reactions that start with a mixture of reactants and products, we first determine reaction direction by comparing Q and K to decide on the sign of x. 

The relationship between Q and signals the direction of a chemical reaction. The free energy change, A G, also signals the direction of a chemical reaction. These two criteria can be compared ... 

In Example 17-4 we calculate the value for Q and compare it with the known value of to predict the direction of the reaction that leads to equilibrium. 

When more of any reactant or product is added to the system, the value of Q changes, so it no longer matches K, and the reaction is no longer at equilibrium. The stress due to the added substance is relieved by shifting the equilibrium in the direction that consumes some of the added substance, moving the value of Q back toward K. Let us compare the mass action expressions for Q and K. If more A or B is added, then Q < K, and the forward reaction occurs more rapidly and to a greater extent than the reverse reaction until equilibrium is reestablished. If more C or D is added, Q > K, and the reverse reaction occurs more rapidly and to a greater extent until equilibrium is reestablished. 

The value of Q in relation to K serves as an index how the composition of the reaction system compares to that of the equilibrium state, and thus it indicates the direction in which any net reaction must proceed. For example, if we combine the two reactants A and B at concentrations of 1 mol L-1 each, the value of Q will indeterminately large(l/0). If instead our mixture consists only of the two products C and D, Q = 0-s-l = 0. It is easy to see (by simply application of the LeChatelier principle) that the ratio of Q/K immediately tells us whether, and in which direction, a net reaction will occur as the system moves toward its equilibrium state ... 

Q. This finding eliminates a truly rapid equilibrium random mechanism, for which k and k must be much smaller than fc 4, k-i, k, and k-2, since the two exchange rates must then be equal. In fact, the differences between the two exchange rates show that the dissociation of A and/or P from the ternary complexes must be slow compared with that of B and/or Q, and also slow relative to the interconversions of the ternary complexes (32). This means that in at least one direction of reaction the dissociation of products in the overall reaction is essentially ordered for all these enzymes, the coenzymes dissociating last, as in the preferred pathway mechanism (Section I,B,4). With malate, lactate, and liver alcohol dehydrogenases, the NAD/NADH exchange rate increased to a... 

We compare the values of Q and K to determine the direction in which a reaction wiii proceed toward equilibrium. 

In the problems we ve worked so far, the direction of the reaction was obvious with only reactants present at the start, the reaction had to go toward products. Thus, in the reaction tables, we knew that the unknown change in reactant concentration had a negative sign (—x) and the change in product concentration had a positive sign (+x). Suppose, however, we start with a mixture of reactants and products. Whenever the reaction direction is not obvious, we first compare the value of Q with K to find the direction in which the reaction proceeds to reach equilibrium. This tells us the sign of x, the unknown change in concentration. (In order to focus on this idea, the next sample problem eliminates the need for the quadratic formula.)... 

Plan (a) To find the direction, we convert the given initial amounts and volume (0.250 L) to concentrations, calculate Q, and compare it with K. (b) Based on the results from (a), we determine the sign of each concentration change for the reaction table and then use the known [CH4] at equilibrium (5.56 M) to determine x and the other equilibrium concentrations. 

In most equilibrium problems, we use quantities (concentrations or pressures) of reactants and products to find K, or we use K to find quantities. We use a reaction table to summarize the initial quantities, how they change, and the equilibrium quantities. When K is small and the initial quantity of reactant is large, we assume the unknown change in the quantity (x) is so much smaller than the initial quantity that it can be neglected. If this assumption is not justified (that is, if the error is greater than 5%), we use the quadratic formula to find x. To determine reaction direction, we compare the values of Q and K. 

In Chapter 17, we compared the values of Q and K to see if a reaction had reached equilibrium and, if not, in which net direction it would move until it did. In this discussion, we use the same approach to see if a precipitate will form and, if not, what changes in the concentrations of the component ions will cause it to do so. As you know, Q p = Kgp when the solution is saturated. If Qgp is greater than Kgp, the solution is momentarily supersaturated, and some solid precipitates until the remaining solution becomes saturated (Qgp = Kgp). If Qgp is less than Kgp, the solution is unsaturated, and no precipitate forms at that temperature (more solid can dissolve). To summarize. 

The sign of AG allows us to predict reaction direction, but you already know that it is not the only way to do so. In Chapter 17, we predicted direction by comparing the values of the reaction quotient (Q) and the equilibrium constant (K). Recall that... 

Thus, we can use the reaction quotient (Q) and the equilibrium constant (K) as guides to help us understand reaction equilibria. For reactions that are not at equilibrium, we can compare Q versus K to determine the direction a reaction should proceed in order to restore its equilibrium. The key points of our discussion of Q and K are summarized in Table 2.2. 

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