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Reaction Kinetics and Mass Action · Tutorial 36 of 799

Equilibrium Constants From Rate Constants

Learn when the ratio of forward to reverse rate constants gives an equilibrium constant, and use that relationship with mass balance to calculate equilibrium concentrations.

Beginner 10 min read

What You'll Learn

  • Derive the equilibrium concentration quotient from equality of forward and reverse rates.
  • Distinguish first-order and higher-order rate-constant units and equilibrium-constant units.
  • Calculate equilibrium concentrations using the rate-constant ratio and conserved totals.
  • Check candidate solutions for nonnegative concentrations, mass balance, and quotient consistency.
  • Recognise when the reversible elementary-step relationship does not apply directly.

From Reversible Rates to an Equilibrium Constant

A reversible reaction can continue in both directions even when its concentrations no longer change. At equilibrium, the forward and reverse rates of that reaction are equal. For a reversible elementary step, this equality links the equilibrium concentration ratio to the forward and reverse rate constants. That link is useful in a QSP model because it allows equilibrium composition to be calculated from kinetic parameters—provided the reaction and system satisfy the relevant assumptions.

Consider the elementary reaction \(A \rightleftharpoons B\). Let \(k_f\) denote its forward rate constant and \(k_r\) its reverse rate constant. The forward rate is \(k_f[A]\), and the reverse rate is \(k_r[B]\), where brackets denote concentration. The net rate of formation of \(B\) is the forward rate minus the reverse rate.

Definition: For a reversible elementary reaction, the concentration equilibrium constant is the equilibrium concentration quotient. For \(A \rightleftharpoons B\), \(K_c=[B]_{\mathrm{eq}}/[A]_{\mathrm{eq}}\). For \(A+B \rightleftharpoons C\), \(K_c=[C]_{\mathrm{eq}}/([A]_{\mathrm{eq}}[B]_{\mathrm{eq}})\). More generally, for \(\nu_A A+\nu_B B \rightleftharpoons \nu_C C\), it is \(K_c=[C]_{\mathrm{eq}}^{\nu_C}/([A]_{\mathrm{eq}}^{\nu_A}[B]_{\mathrm{eq}}^{\nu_B})\). The subscripts indicate equilibrium concentrations.

These quotients follow from mass-action rate laws for the elementary forward and reverse steps. In this tutorial, \(K_c\) refers to the concentration quotient; depending on the reaction stoichiometry, it can have units. A thermodynamic equilibrium constant defined using dimensionless activities is a different convention and should not be confused with a dimensionful concentration quotient.

Deriving the Rate-Constant Ratio

The derivation requires a reversible elementary step whose forward and reverse rates follow mass action, with constant rate constants under the conditions considered. It also assumes the step can reach equilibrium and that other reactions or external inputs do not keep this step from having equal forward and reverse rates at equilibrium. For an isolated step, the net rate is zero at equilibrium—not because both directional rates have vanished, but because they are equal.

$$ v_{\mathrm{net}} = k_f[A]^{\nu_A}[B]^{\nu_B} - k_r[C]^{\nu_C} $$

At equilibrium, \(v_{\mathrm{net}}=0\). Setting the forward and reverse terms equal, then dividing by the nonzero rate constant \(k_r\) and the reactant concentrations, gives the concentration quotient.

$$ k_f[A]_{\mathrm{eq}}^{\nu_A}[B]_{\mathrm{eq}}^{\nu_B} = k_r[C]_{\mathrm{eq}}^{\nu_C} \quad\Longrightarrow\quad K_c = \frac{[C]_{\mathrm{eq}}^{\nu_C}} {[A]_{\mathrm{eq}}^{\nu_A}[B]_{\mathrm{eq}}^{\nu_B}} = \frac{k_f}{k_r} $$

Thus the rate-constant ratio determines the equilibrium quotient for this reversible elementary step. For \(A\rightleftharpoons B\), \(K_c=k_f/k_r\). For \(A+B\rightleftharpoons C\), the same ratio equals \([C]_{\mathrm{eq}}/([A]_{\mathrm{eq}}[B]_{\mathrm{eq}})\). The units must agree: for \(A+B\rightleftharpoons C\), \(k_f\) has units of concentration\(^{-1}\) time\(^{-1}\), \(k_r\) has units of time\(^{-1}\), and \(K_c\) has units of concentration\(^{-1}\).

The ratio alone specifies a concentration relationship, not the absolute concentrations. To calculate those concentrations, combine the quotient with the initial amounts and any conserved totals. The physically relevant solution must have nonnegative concentrations and satisfy those conservation constraints.

Worked Example: A First-Order Reversible Step

Suppose \(A\rightleftharpoons B\) has \(k_f=0.30\ \mathrm{h}^{-1}\) and \(k_r=0.10\ \mathrm{h}^{-1}\). Initially, the closed system contains \(12\ \mathrm{\mu M}\) total, with no other reactions. Find the equilibrium concentrations.

For this first-order step, both rate constants have units of inverse time, so their ratio is dimensionless. The equilibrium constant is

$$ K_c=\frac{k_f}{k_r} =\frac{0.30\ \mathrm{h}^{-1}}{0.10\ \mathrm{h}^{-1}} =3 $$

The quotient says \([B]_{\mathrm{eq}}=3[A]_{\mathrm{eq}}\). Conservation of the total concentration gives \([A]_{\mathrm{eq}}+[B]_{\mathrm{eq}}=12\ \mathrm{\mu M}\). Substitution yields

$$ [A]_{\mathrm{eq}}+3[A]_{\mathrm{eq}}=12\ \mathrm{\mu M} \quad\Longrightarrow\quad [A]_{\mathrm{eq}}=3.00\ \mathrm{\mu M}, \qquad [B]_{\mathrm{eq}}=9.00\ \mathrm{\mu M} $$

The concentrations are nonnegative, sum to the conserved \(12\ \mathrm{\mu M}\), and have quotient \(9.00/3.00=3\), as required.

Using Conservation to Find Equilibrium Concentrations

For reactions involving multiple reactants, the equilibrium quotient may be nonlinear in the concentrations. A practical method is to express all concentrations using one reaction-progress variable, apply the rate-constant ratio, and solve the resulting equation. Conservation laws then constrain which mathematical roots are physically possible.

Worked Example: Association of Two Reactants

Consider \(A+B\rightleftharpoons C\), with \(k_f=0.10\ \mathrm{\mu M}^{-1}\mathrm{min}^{-1}\) and \(k_r=0.20\ \mathrm{min}^{-1}\). Initially, \([A]=10\ \mathrm{\mu M}\), \([B]=8\ \mathrm{\mu M}\), and \([C]=0\). Assume a closed system with this reaction as the only process. Calculate the equilibrium concentrations.

The concentration equilibrium constant is

$$ K_c=\frac{k_f}{k_r} =\frac{0.10\ \mathrm{\mu M}^{-1}\mathrm{min}^{-1}} {0.20\ \mathrm{min}^{-1}} =0.500\ \mathrm{\mu M}^{-1} $$

Let \(x\ \mathrm{\mu M}\) of \(C\) form. The reaction consumes one unit of \(A\) and one unit of \(B\) per unit of \(C\), so the concentrations are \([A]=10-x\), \([B]=8-x\), and \([C]=x\), all in \(\mathrm{\mu M}\). The constraints \(0\leq x\leq 8\) ensure no concentration is negative. Substitute these expressions into the equilibrium quotient:

$$ \frac{x}{(10-x)(8-x)}=0.500 \quad\Longrightarrow\quad x=0.500(10-x)(8-x) $$

Expanding and solving,

$$ x^2-20x+80=0 \quad\Longrightarrow\quad x=10\pm\sqrt{20} $$

The roots are approximately \(5.528\) and \(14.47\ \mathrm{\mu M}\). Only \(x=5.528\ \mathrm{\mu M}\) lies in the physically allowed range \(0\leq x\leq 8\). Therefore,

$$ [A]_{\mathrm{eq}}=4.472\ \mathrm{\mu M},\qquad [B]_{\mathrm{eq}}=2.472\ \mathrm{\mu M},\qquad [C]_{\mathrm{eq}}=5.528\ \mathrm{\mu M} $$

As a check, \(4.472+5.528=10.000\ \mathrm{\mu M}\) of \(A\)-units and \(2.472+5.528=8.000\ \mathrm{\mu M}\) of \(B\)-units. Also, \(5.528/(4.472\times2.472)\approx0.500\ \mathrm{\mu M}^{-1}\), consistent with \(K_c\).

Worked Example: A Dimerisation Step

For the elementary step \(2A\rightleftharpoons B\), let \(k_f=0.040\ \mathrm{\mu M}^{-1}\mathrm{min}^{-1}\) and \(k_r=0.20\ \mathrm{min}^{-1}\). The conserved total concentration of \(A\)-units is \(10\ \mathrm{\mu M}\), so \([A]+2[B]=10\ \mathrm{\mu M}\). Find the equilibrium concentrations.

The forward rate is \(k_f[A]^2\), while the reverse rate is \(k_r[B]\). Therefore,

$$ K_c=\frac{[B]_{\mathrm{eq}}}{[A]_{\mathrm{eq}}^2} =\frac{k_f}{k_r} =\frac{0.040}{0.20} =0.200\ \mathrm{\mu M}^{-1} $$

Write \(a=[A]_{\mathrm{eq}}\) and \(b=[B]_{\mathrm{eq}}\). The quotient gives \(b=0.200a^2\), and conservation gives \(a+2b=10\). Combining them,

$$ 0.400a^2+a-10=0 \quad\Longrightarrow\quad a=\frac{-1\pm\sqrt{17}}{0.800} $$

The negative root is not a possible concentration. The positive root gives \(a=3.904\ \mathrm{\mu M}\), and then \(b=0.200(3.904)^2=3.048\ \mathrm{\mu M}\). The conservation check is \(3.904+2(3.048)=10.000\ \mathrm{\mu M}\), and the quotient check is \(3.048/(3.904)^2\approx0.200\ \mathrm{\mu M}^{-1}\).

Checks, Alternatives, and Limitations

A calculated equilibrium composition should satisfy three checks: the concentrations are nonnegative, all conserved totals are respected, and substitution into the concentration quotient returns \(k_f/k_r\). If the model predicts a negative concentration or violates a total, the selected algebraic root is not physically valid. If no root satisfies the constraints, check the rate law, units, stoichiometry, and conservation equations before interpreting the calculation.

For a single reversible elementary step, solving the quotient and conservation equations is often simpler than simulating the ODEs until concentrations appear to settle. ODE simulation is useful for examining the approach to equilibrium or checking a calculation over time, but it requires suitable initial conditions and a sufficiently long simulation. Conversely, measured equilibrium concentrations can estimate a quotient, but that quotient alone does not identify \(k_f\) and \(k_r\) separately; many pairs of rates have the same ratio.

The ratio result is conditional on the elementary mass-action model. If an observed reaction is actually a collection of hidden steps, its apparent rate law need not have the elementary form used in the derivation. Likewise, in a network with additional reactions or ongoing inputs, zero net change in one concentration does not by itself establish that the forward and reverse fluxes of this particular step are equal. The relationship should be applied to the specified reversible elementary step under conditions where its forward and reverse rates balance.

  • Reversing the ratio. Using \(k_r/k_f\) gives the reciprocal of the defined concentration quotient. Check which species are in the numerator before calculating.
  • Ignoring units. For \(A+B\rightleftharpoons C\), \(K_c\) has units of inverse concentration. A unit mismatch can reveal an incorrect rate law or quotient.
  • Keeping every quadratic root. Algebraic roots are candidates, not automatically valid concentrations. Apply nonnegativity and conservation constraints.
  • Equating equilibrium with zero directional rates. The forward and reverse rates can both be positive at equilibrium; it is their difference that is zero.
Key takeaway: For a reversible elementary reaction with mass-action rates, setting the forward and reverse rates equal gives \(K_c=k_f/k_r\). Use this quotient together with conservation laws, then verify units, nonnegative concentrations, and mass balance.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. For \(X\rightleftharpoons Y\), \(k_f=0.12\ \mathrm{min}^{-1}\) and \(k_r=0.04\ \mathrm{min}^{-1}\). If the conserved total is \(16\ \mathrm{\mu M}\), what are the equilibrium concentrations?
  2. For \(M+N\rightleftharpoons P\), what are the units of \(K_c\) if concentration is measured in \(\mathrm{\mu M}\)?
  3. For \(2R\rightleftharpoons S\), write the equilibrium quotient in terms of concentrations and state the mass-balance relation if the total concentration of \(R\)-units is \(T\).
  4. A calculation for \(A+B\rightleftharpoons C\) yields two roots, \(x=4\ \mathrm{\mu M}\) and \(x=11\ \mathrm{\mu M}\), for product formed; the initial concentration of \(B\) is \(7\ \mathrm{\mu M}\). Which root can be used, and why?

Solutions. 1. \(K_c=0.12/0.04=3\), so \([Y]=3[X]\). With \([X]+[Y]=16\ \mathrm{\mu M}\), \([X]=4.00\ \mathrm{\mu M}\) and \([Y]=12.0\ \mathrm{\mu M}\). 2. \(K_c=[P]/([M][N])\), so its units are \(\mathrm{\mu M}^{-1}\). 3. \(K_c=[S]/[R]^2=k_f/k_r\), and conservation is \([R]+2[S]=T\). 4. Use \(x=4\ \mathrm{\mu M}\) only. Product formation cannot exceed the initial \(7\ \mathrm{\mu M}\) of \(B\), so \(x=11\ \mathrm{\mu M}\) would imply a negative remaining concentration of \(B\).

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