To know the partnership between the equilibrium continuous and also the rate constants for the forward and reverse reactions. To create an equilibrium constant expression for any kind of reaction.

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Since an equilibrium state is achieved when the forward reaction rate equals the reverse reaction price, under a offered collection of conditions tbelow need to be a partnership in between the complace of the system at equilibrium and also the kinetics of a reaction (represented by rate constants). We can present this relationship utilizing the decomplace reactivity of (ceN_2O_4) to (ceNO2). Both the forward and also reverse reactions for this system consist of a solitary elementary reactivity, so the reaction rates are as follows:

< extforward rate = k_f onumber>

and

< extreverse rate = k_r^2 onumber>

At equilibrium, the forward rate equates to the reverse price (definition of equilibrium):

< k_f = k_r^2 labelEq3>

so

^2 labelEq4>

The ratio of the rate constants provides us a brand-new continuous, the equilibrium consistent ((K)), which is defined as follows:

Hence tbelow is a fundamental relationship in between muzic-ivan.infoical kinetics and also muzic-ivan.infoical equilibrium: under a given set of problems, the composition of the equilibrium mixture is figured out by the magnitudes of the price constants for the forward and also the reverse reactions.

The equilibrium continuous is equal to the price consistent for the forward reactivity divided by the rate consistent for the reverse reactivity.

Table (PageIndex1) lists the initial and also equilibrium concentrations from five various experiments utilizing the reaction system explained by Equation ( efEq3). At equilibrium the magnitude of the quantity (^2/) is essentially the exact same for all five experiments. In truth, no matter what the initial concentrations of (ceNO2) and (ceN2O4) are, at equilibrium the amount (^2/) will certainly constantly be (6.53 pm 0.03 imes 10^-3) at 25°C, which corresponds to the proportion of the price constants for the forward and reverse reactions. That is, at a offered temperature, the equilibrium continuous for a reaction always has actually the same worth, even though the particular concentrations of the reactants and assets vary depending on their initial concentrations.

Table (PageIndex1): Initial and also Equilibrium Concentrations for (ceNO2:ceN2O4) Mixtures at 25°C Initial ConcentrationsConcentrations at Equilibrium Experiment <(ceN2O4)> (M) <(ceNO2)> (M) <(ceN2O4)> (M) <(ceNO2)> (M) (K = ^2/)
1 0.0500 0.0000 0.0417 0.0165 (6.54 imes 10^-3)
2 0.0000 0.1000 0.0417 0.0165 (6.54 imes 10^-3)
3 0.0750 0.0000 0.0647 0.0206 (6.56 imes 10^-3)
4 0.0000 0.0750 0.0304 0.0141 (6.54 imes 10^-3)
5 0.0250 0.0750 0.0532 0.0186 (6.50 imes 10^-3)

Developing an Equilibrium Constant Expression

In 1864, the Norwegian muzic-ivan.infoists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) very closely measured the compositions of many reaction devices at equilibrium. They found that for any type of reversible reactivity of the basic form

wbelow A and also B are reactants, C and also D are commodities, and a, b, c, and d are the stoichiometric coefficients in the balanced muzic-ivan.infoical equation for the reaction, the proportion of the product of the equilibrium concentrations of the assets (raised to their coefficients in the balanced muzic-ivan.infoical equation) to the product of the equilibrium concentrations of the reactants (increased to their coefficients in the balanced muzic-ivan.infoical equation) is constantly a consistent under a offered set of problems. This partnership is well-known as the legislation of mass activity (or legislation of muzic-ivan.infoical equilibrium) and have the right to be proclaimed as follows:

^c^d^a^b labelEq7>

wbelow (K) is the equilibrium constant for the reaction. Equation ( efEq6) is called the equilibrium equation, and the right side of Equation ( efEq7) is referred to as the equilibrium constant expression. The connection shown in Equation ( efEq7) is true for any pair of opposing reactions regardmuch less of the mechanism of the reactivity or the variety of procedures in the system.

The equilibrium continuous can vary over a wide array of worths. The values of (K) shown in Table (PageIndex2), for example, differ by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and also reactants are in the denominator, worths of K greater than (10^3) suggest a strong tendency for reactants to develop assets. In this case, muzic-ivan.infoists say that equilibrium lies to the ideal as written, favoring the formation of commodities. An example is the reactivity between (H_2) and (Cl_2) to produce (HCl), which has an equilibrium consistent of (1.6 imes 10^33) at 300 K. Due to the fact that (H_2) is a great reductant and (Cl_2) is a good oxidant, the reaction proceeds basically to completion. In contrast, values of (K) less than (10^-3) suggest that the proportion of commodities to reactants at equilibrium is extremely small. That is, reactants execute not tfinish to form assets conveniently, and the equilibrium lies to the left as composed, favoring the formation of reactants.

Table (PageIndex2): Equilibrium Constants for Schosen Reactions* Reaction Temperature (K) Equilibrium Constant (K) *Equilibrium constants vary with temperature. The K worths displayed are for devices at the suggested temperatures.
(S_(s)+O_2(g) ightleftharpoons SO_2(g)) 300 (4.4 imes 10^53)
(2H_2(g)+O_2(g) ightleftharpoons 2H2O_(g)) 500 (2.4 imes 10^47)
(H_2(g)+Cl_2(g) ightleftharpoons 2HCl_(g)) 300 (1.6 imes 10^33)
(H_2(g)+Br_2(g) ightleftharpoons 2HBr_(g)) 300 (4.1 imes 10^18)
(2NO_(g)+O_2(g) ightleftharpoons 2NO_2(g)) 300 (4.2 imes 10^13)
(3H_2(g)+N_2(g) ightleftharpoons 2NH_3(g)) 300 (2.7 imes 10^8)
(H_2(g)+D_2(g) ightleftharpoons 2HD_(g)) 100 (1.92)
(H_2(g)+I_2(g) ightleftharpoons 2HI_(g)) 300 (2.9 imes 10^-1)
(I_2(g) ightleftharpoons 2I_(g)) 800 (4.6 imes 10^-7)
(Br_2(g) ightleftharpoons 2Br_(g)) 1000 (4.0 imes 10^-7)
(Cl_2(g) ightleftharpoons 2Cl_(g)) 1000 (1.8 imes 10^-9)
(F_2(g) ightleftharpoons 2F_(g)) 500 (7.4 imes 10^-13)

Effective vs. True Concentrations

You will additionally alert in Table (PageIndex2) that equilibrium constants have no devices, also though Equation ( efEq7) says that the units of concentration might not always cancel because the exponents might vary. In reality, equilibrium constants are calculated making use of “efficient concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a traditional state of 1 M. As displayed in Equation ( efEq8), the devices of concentration cancel, which provides (K) unitmuch less as well:

< dfrac_measured_standard; state=dfraccancelMcancelM = dfraccancelfracmolLcancelfracmolL labelEq8>

Due to the fact that equilibrium constants are calculated utilizing “effective concentrations” relative to a standard state of 1 M, values of K are unitless.


Many kind of reactions have actually equilibrium constants in between 1000 and also 0.001 ((10^3 ge K ge 10^-3)), neither extremely large nor very tiny. At equilibrium, these units tfinish to contain considerable quantities of both products and reactants, indicating that there is not a solid tendency to develop either commodities from reactants or reactants from commodities. An instance of this kind of mechanism is the reactivity of gaseous hydrogen and also deuterium, a component of high-stcapacity fiber-optic light resources provided in sea research studies, to develop (ceHD):

The equilibrium consistent expression for this reactivity is

^2 onumber>

via (K) varying in between 1.9 and 4 over a vast temperature range (100–1000 K). Hence an equilibrium mixture of (H_2), (D_2), and also (HD) contains considerable concentrations of both product and reactants.

Figure (PageIndex3) summarizes the relationship between the magnitude of K and the relative concentrations of reactants and commodities at equilibrium for a basic reaction, written as reactants ( ightleftharpoons) products. Because there is a direct partnership between the kinetics of a reactivity and the equilibrium concentrations of commodities and reactants (Equations ( efEq8) and ( efEq7)), as soon as (k_f gg k_r), (K) is a large number, and also the concentration of products at equilibrium preconquer. This synchronizes to an essentially irreversible reactivity. Conversely, as soon as (k_f ll k_r), (K) is an extremely little number, and also the reaction produces almost no products as composed. Equipment for which (k_f ≈ k_r) have considerable concentrations of both reactants and also assets at equilibrium.

*
Figure (PageIndex3): The Relationship between the Complace of the Mixture at Equilibrium and the Magnitude of the Equilibrium Constant. The larger the K, the farther the reactivity proceeds to the appropriate prior to equilibrium is reached, and the greater the proportion of commodities to reactants at equilibrium.

A big worth of the equilibrium continuous (K) suggests that commodities preovercome at equilibrium; a little value implies that reactants predominate at equilibrium.






Variations in the Form of the Equilibrium Constant Expression

Since equilibrium can be approached from either direction in a muzic-ivan.infoical reaction, the equilibrium constant expression and for this reason the magnitude of the equilibrium continuous depend on the develop in which the muzic-ivan.infoical reaction is composed. For example, if we create the reaction described in Equation ( efEq6) in reverse, we attain the following:

The matching equilibrium consistent (K′) is as follows:

^a^b^c^d labelEq11>

This expression is the inverse of the expression for the original equilibrium consistent, so (K′ = 1/K). That is, as soon as we create a reaction in the reverse direction, the equilibrium constant expression is inverted. For circumstances, the equilibrium continuous for the reaction (ceN2O4 2NO2) is as follows:

^2 labelEq12>

however for the oppowebsite reactivity, (2 NO_2 ightleftharpoons N_2O_4), the equilibrium continuous K′ is offered by the inverse expression:

^2 labelEq13>

Consider one more instance, the development of water:

Since (ceH2) is an excellent reductant and (ceO2) is a great oxidant, this reactivity has actually a very huge equilibrium consistent ((K = 2.4 imes 10^47) at 500 K). Consequently, the equilibrium consistent for the reverse reaction, the decomplace of water to develop (ceO2) and also (ceH2), is extremely small: (K′ = 1/K = 1/(2.4 imes 10^47) = 4.2 imes 10^-48). As said by the very little equilibrium consistent, and fortunately for life as we recognize it, a substantial amount of energy is indeed required to dissociate water into (ceH2) and (ceO2).

The equilibrium continuous for a reactivity composed in reverse is the inverse of the equilibrium consistent for the reaction as written initially.

Writing an equation in various however muzic-ivan.infoically indistinguishable forms additionally causes both the equilibrium consistent expression and the magnitude of the equilibrium continuous to be various. For example, we could compose the equation for the reaction

as

via the equilibrium constant K″ is as follows:

< K′′=dfrac^1/2 labelEq14>

The worths for K′ (Equation ( efEq13)) and also K″ are connected as follows:

< K′′=(K")^1/2=sqrtK" labelEq15>

In basic, if all the coefficients in a balanced muzic-ivan.infoical equation were consequently multiplied by (n), then the new equilibrium continuous is the original equilibrium constant raised to the (n^th) power.




Equilibrium Constant Expressions for Systems that Contain Gases

For reactions that involve species in solution, the concentrations offered in equilibrium calculations are commonly expressed in moles/liter. For gases, yet, the concentrations are typically expressed in terms of partial pressures quite than molarity, where the typical state is 1 atm of push. The symbol (K_p) is used to represent equilibrium constants calculated from partial pressures. For the general reaction (aA+bB ightleftharpoons cC+dD), in which all the components are gases, the equilibrium consistent expression can be created as the ratio of the partial pressures of the assets and also reactants (each elevated to its coreliable in the muzic-ivan.infoical equation):

Therefore (K_p) for the decomplace of (N_2O_4) (Equation 15.1) is as follows:

Like (K), (K_p) is a unitless amount bereason the amount that is actually used to calculate it is an “reliable press,” the proportion of the measured push to a conventional state of 1 bar (about 1 atm), which produces a unitless amount. The “efficient pressure” is referred to as the fugacity, simply as activity is the effective concentration.

Since partial pressures are usually expressed in environments or mmHg, the molar concentration of a gas and also its partial push carry out not have the same numerical worth. Consequently, the numerical worths of (K) and also (K_p) are typically various. They are, but, associated by the appropriate gas continuous ((R)) and the absolute temperature ((T)):

where (K) is the equilibrium constant expressed in devices of concentration and (Δn) is the difference in between the numbers of moles of gaseous products and gaseous reactants ((n_p − n_r)). The temperature is expressed as the absolute temperature in Kelvin. According to Equation ( efEq18), (K_p = K) just if the moles of gaseous assets and gaseous reactants are the very same (i.e., (Δn = 0)). For the decomposition of (N_2O_4), tright here are 2 mol of gaseous product and also 1 mol of gaseous reactant, so (Δn = 1). Therefore, for this reactivity,




Equilibrium Constant Expressions for the Sums of Reactions

muzic-ivan.infoists frequently must understand the equilibrium continuous for a reactivity that has not been formerly studied. In such cases, the preferred reactivity have the right to often be written as the amount of other reactions for which the equilibrium constants are known. The equilibrium constant for the unrecognized reaction can then be calculated from the tabulated worths for the other reactions.

To illustrate this procedure, let’s think about the reactivity of (ceN2) via (ceO2) to offer (ceNO2). This reactivity is an important resource of the (ceNO2) that provides metropolitan smog its typical brown color. The reactivity typically occurs in two distinctive procedures. In the first reactivity (step 1), (ceN2) reacts through (ceO2) at the high temperatures inside an internal combustion engine to offer (ceNO). The released (ceNO) then reacts with additional (ceO2) to give (ceNO2) (action 2). The equilibrium consistent for each reaction at 100°C is additionally offered.

(ceN2(g) + O2(g) 2NO(g);; K_1=2.0 imes 10^-25 labelaction 1)

(ce2NO(g) + O2(g) 2NO2(g);;;K_2=6.4 imes 10^9 labelaction 2)

Summing reactions (action 1) and also (action 2) provides the all at once reactivity of (N_2) through (O_2):

(ceN2(g) + 2O2(g) 2NO2(g) ;;;K_3=? labelas a whole reactivity 3)

The equilibrium consistent expressions for the reactions are as follows:

^2;;; K_2=dfrac^2^2;;; K_3=dfrac^2^2>

What is the connection between (K_1), (K_2), and also (K_3), all at 100°C? The expression for (K_1) has actually (^2) in the numerator, the expression for (K_2) has actually (^2) in the denominator, and also (^2) does not appear in the expression for (K_3). Multiplying (K_1) by (K_2) and canceling the (^2) terms,

< K_1K_2=dfraccancel^2 imes dfrac^2cancel^2=dfrac^2^2=K_3>

Thus the product of the equilibrium continuous expressions for (K_1) and also (K_2) is the same as the equilibrium continuous expression for (K_3):

The equilibrium constant for a reactivity that is the amount of two or even more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recontact that according to Hess’s Law, (ΔH) for the amount of two or even more reactions is the sum of the ΔH worths for the individual reactions.

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To determine (K) for a reactivity that is the sum of two or more reactions, include the reactions but multiply the equilibrium constants.