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Chemistry Notebook

Reversible Reaction Kinetics and Relaxation to Equilibrium

Many chemical reactions do not proceed only in one direction. As products form, they may react to regenerate the original reactants. The observed change in concentration is then determined by the difference between the forward and reverse reaction rates.

Consider the reversible first-order reaction

A⇌k1k−1B.A \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} B.

The forward rate is

vf=k1[A],v_{\mathrm f}=k_1[A],

and the reverse rate is

vr=k−1[B].v_{\mathrm r}=k_{-1}[B].

The net rate of formation of BB is therefore

d[B]dt=k1[A]−k−1[B].\boxed{ \frac{d[B]}{dt} = k_1[A]-k_{-1}[B]. }

Similarly,

d[A]dt=−k1[A]+k−1[B].\frac{d[A]}{dt} = -k_1[A]+k_{-1}[B].

At equilibrium, the concentrations no longer change with time, so

d[B]dt=0.\frac{d[B]}{dt}=0.

Therefore,

k1[A]eq=k−1[B]eq.k_1[A]_{\mathrm{eq}} = k_{-1}[B]_{\mathrm{eq}}.

Rearranging,

[B]eq[A]eq=k1k−1.\boxed{ \frac{[B]_{\mathrm{eq}}}{[A]_{\mathrm{eq}}} = \frac{k_1}{k_{-1}}. }

For this elementary reversible reaction,

K=k1k−1.\boxed{ K=\frac{k_1}{k_{-1}}. }

Equilibrium therefore does not mean that the forward and reverse reactions have stopped. Both reactions continue, but their rates are equal.

Concentration as a Function of Time

Section titled “Concentration as a Function of Time”

Suppose the total concentration of AA and BB remains constant:

[A]+[B]=C.[A]+[B]=C.

Thus,

[A]=C−[B].[A]=C-[B].

Substituting this into

d[B]dt=k1[A]−k−1[B]\frac{d[B]}{dt} = k_1[A]-k_{-1}[B]

gives

d[B]dt=k1(C−[B])−k−1[B].\frac{d[B]}{dt} = k_1(C-[B])-k_{-1}[B].

Expanding,

d[B]dt=k1C−k1[B]−k−1[B].\frac{d[B]}{dt} = k_1C-k_1[B]-k_{-1}[B].

Therefore,

d[B]dt=k1C−(k1+k−1)[B].\frac{d[B]}{dt} = k_1C-(k_1+k_{-1})[B].

At equilibrium,

0=k1C−(k1+k−1)[B]eq.0 = k_1C-(k_1+k_{-1})[B]_{\mathrm{eq}}.

Hence,

[B]eq=k1Ck1+k−1.[B]_{\mathrm{eq}} = \frac{k_1C}{k_1+k_{-1}}.

Similarly,

[A]eq=k−1Ck1+k−1.[A]_{\mathrm{eq}} = \frac{k_{-1}C}{k_1+k_{-1}}.

The differential equation can now be written in terms of the distance from equilibrium. Since

k1C=(k1+k−1)[B]eq,k_1C = (k_1+k_{-1})[B]_{\mathrm{eq}},

we have

d[B]dt=(k1+k−1)[B]eq−(k1+k−1)[B].\frac{d[B]}{dt} = (k_1+k_{-1})[B]_{\mathrm{eq}} - (k_1+k_{-1})[B].

Therefore,

d[B]dt=−(k1+k−1)([B]−[B]eq).\frac{d[B]}{dt} = -(k_1+k_{-1}) \left( [B]-[B]_{\mathrm{eq}} \right).

Define

y=[B]−[B]eq.y=[B]-[B]_{\mathrm{eq}}.

Because [B]eq[B]_{\mathrm{eq}} is constant,

dy=d[B].dy=d[B].

The differential equation becomes

dydt=−(k1+k−1)y.\frac{dy}{dt} = -(k_1+k_{-1})y.

Separate the variables:

dyy=−(k1+k−1)dt.\frac{dy}{y} = -(k_1+k_{-1})dt.

Integrating from y0y_0 at t=0t=0 to yy at time tt,

∫y0ydyy=−(k1+k−1)∫0tdt.\int_{y_0}^{y}\frac{dy}{y} = -(k_1+k_{-1}) \int_0^t dt.

This gives

ln⁡yy0=−(k1+k−1)t.\ln\frac{y}{y_0} = -(k_1+k_{-1})t.

Exponentiating,

y=y0e−(k1+k−1)t.y = y_0e^{-(k_1+k_{-1})t}.

Substituting back,

[B]−[B]eq=([B]0−[B]eq)e−(k1+k−1)t.\boxed{ [B]-[B]_{\mathrm{eq}} = \left( [B]_0-[B]_{\mathrm{eq}} \right) e^{-(k_1+k_{-1})t}. }

or

[B]=[B]eq+([B]0−[B]eq)e−(k1+k−1)t.\boxed{ [B] = [B]_{\mathrm{eq}} + \left( [B]_0-[B]_{\mathrm{eq}} \right) e^{-(k_1+k_{-1})t}. }

The same result can be written for AA:

[A]=[A]eq+([A]0−[A]eq)e−(k1+k−1)t.\boxed{ [A] = [A]_{\mathrm{eq}} + \left( [A]_0-[A]_{\mathrm{eq}} \right) e^{-(k_1+k_{-1})t}. }

The important difference from an irreversible first-order reaction is the concentration approached at long times.

For

A→B,A\rightarrow B,

the concentration of AA approaches zero:

[A]→0.[A]\rightarrow0.

For

A⇌B,A\rightleftharpoons B,

the concentration approaches its equilibrium value:

[A]→[A]eq.[A]\rightarrow[A]_{\mathrm{eq}}.

The quantity that decays exponentially is therefore not the concentration itself, but its deviation from equilibrium.

Relaxation Time

Section titled “Relaxation Time”

From

[B]−[B]eq=([B]0−[B]eq)e−(k1+k−1)t,[B]-[B]_{\mathrm{eq}} = \left( [B]_0-[B]_{\mathrm{eq}} \right) e^{-(k_1+k_{-1})t},

the characteristic relaxation time is

τ=1k1+k−1.\boxed{ \tau = \frac{1}{k_1+k_{-1}}. }

The concentration can therefore be written as

[B]−[B]eq=([B]0−[B]eq)e−t/τ.[B]-[B]_{\mathrm{eq}} = \left( [B]_0-[B]_{\mathrm{eq}} \right)e^{-t/\tau}.

After one relaxation time,

t=τ,t=\tau,

so

[B]−[B]eq[B]0−[B]eq=e−1.\frac{[B]-[B]_{\mathrm{eq}}} {[B]_0-[B]_{\mathrm{eq}}} = e^{-1}.

Thus the deviation from equilibrium has fallen to about 37%37\% of its initial value.

The half-time for relaxation is obtained from

[B]−[B]eq[B]0−[B]eq=12.\frac{[B]-[B]_{\mathrm{eq}}} {[B]_0-[B]_{\mathrm{eq}}} = \frac12.

Therefore,

12=e−(k1+k−1)t1/2.\frac12 = e^{-(k_1+k_{-1})t_{1/2}}.

Taking logarithms,

ln⁡12=−(k1+k−1)t1/2.\ln\frac12 = -(k_1+k_{-1})t_{1/2}.

Hence,

t1/2=ln⁡2k1+k−1.\boxed{ t_{1/2} = \frac{\ln2}{k_1+k_{-1}}. }

Determining the Forward and Reverse Rate Constants

Section titled “Determining the Forward and Reverse Rate Constants”

For the reversible first-order reaction,

A⇌B,A\rightleftharpoons B,

equilibrium measurements give

K=k1k−1,K=\frac{k_1}{k_{-1}},

while kinetic measurements of the relaxation give

1τ=k1+k−1.\frac{1}{\tau}=k_1+k_{-1}.

These two relations can be combined to determine the individual rate constants.

From

k1=Kk−1,k_1=Kk_{-1},

substitution into the relaxation equation gives

1τ=Kk−1+k−1.\frac{1}{\tau} = Kk_{-1}+k_{-1}.

Therefore,

1τ=(K+1)k−1,\frac{1}{\tau} = (K+1)k_{-1},

and

k−1=1τ(K+1).\boxed{ k_{-1} = \frac{1}{\tau(K+1)}. }

Since

k1=Kk−1,k_1=Kk_{-1},

we obtain

k1=Kτ(K+1).\boxed{ k_1 = \frac{K}{\tau(K+1)}. }

Thus equilibrium data provide the ratio of the forward and reverse rate constants, while relaxation data provide their sum.

Reversible Reactions with Different Rate Laws

Section titled “Reversible Reactions with Different Rate Laws”

The simple exponential result above applies to a reversible first-order system. More complicated reversible reactions do not necessarily give the same form.

Consider

A+B⇌k1k−1C.A+B \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} C.

If both directions are elementary,

vf=k1[A][B]v_{\mathrm f} = k_1[A][B]

and

vr=k−1[C].v_{\mathrm r} = k_{-1}[C].

The net rate of formation of CC is

d[C]dt=k1[A][B]−k−1[C].\boxed{ \frac{d[C]}{dt} = k_1[A][B]-k_{-1}[C]. }

At equilibrium,

k1[A]eq[B]eq=k−1[C]eq.k_1[A]_{\mathrm{eq}}[B]_{\mathrm{eq}} = k_{-1}[C]_{\mathrm{eq}}.

Therefore,

[C]eq[A]eq[B]eq=k1k−1.\boxed{ \frac{[C]_{\mathrm{eq}}} {[A]_{\mathrm{eq}}[B]_{\mathrm{eq}}} = \frac{k_1}{k_{-1}}. }

For this elementary mechanism,

Kc=k1k−1.K_c=\frac{k_1}{k_{-1}}.

The time dependence is more complicated than for the first-order reversible reaction because the differential equation contains the product [A][B][A][B].

Suppose the initial concentrations are

[A]0=a,[A]_0=a, [B]0=b,[B]_0=b,

and

[C]0=c.[C]_0=c.

If the net extent of the forward reaction at time tt is xx, then

[A]=a−x,[A]=a-x, [B]=b−x,[B]=b-x,

and

[C]=c+x.[C]=c+x.

Therefore,

dxdt=k1(a−x)(b−x)−k−1(c+x).\frac{dx}{dt} = k_1(a-x)(b-x) - k_{-1}(c+x).

Unlike the reversible first-order case, this equation is nonlinear in xx.

Relaxation Near Equilibrium

Section titled “Relaxation Near Equilibrium”

Even when the full rate equation is nonlinear, the behavior close to equilibrium can often be simplified.

Consider again

A+B⇌C.A+B\rightleftharpoons C.

Let the equilibrium concentrations be

[A]eq=Ae,[A]_{\mathrm{eq}}=A_e, [B]eq=Be,[B]_{\mathrm{eq}}=B_e,

and

[C]eq=Ce.[C]_{\mathrm{eq}}=C_e.

Suppose the system is displaced slightly from equilibrium by an amount xx. Then

[A]=Ae−x,[A]=A_e-x, [B]=Be−x,[B]=B_e-x,

and

[C]=Ce+x.[C]=C_e+x.

The net rate is

dxdt=k1(Ae−x)(Be−x)−k−1(Ce+x).\frac{dx}{dt} = k_1(A_e-x)(B_e-x) - k_{-1}(C_e+x).

Expanding,

dxdt=k1AeBe−k1Aex−k1Bex+k1x2−k−1Ce−k−1x.\frac{dx}{dt} = k_1A_eB_e -k_1A_ex -k_1B_ex +k_1x^2 -k_{-1}C_e -k_{-1}x.

At equilibrium,

k1AeBe=k−1Ce,k_1A_eB_e = k_{-1}C_e,

so these terms cancel:

dxdt=−[k1(Ae+Be)+k−1]x+k1x2.\frac{dx}{dt} = -\left[ k_1(A_e+B_e)+k_{-1} \right]x + k_1x^2.

For a small displacement from equilibrium,

x2≪x,x^2\ll x,

so the quadratic term can be neglected:

dxdt≈−[k1(Ae+Be)+k−1]x.\frac{dx}{dt} \approx -\left[ k_1(A_e+B_e)+k_{-1} \right]x.

Separate the variables:

dxx=−[k1(Ae+Be)+k−1]dt.\frac{dx}{x} = -\left[ k_1(A_e+B_e)+k_{-1} \right]dt.

Integrating,

ln⁡xx0=−[k1(Ae+Be)+k−1]t.\ln\frac{x}{x_0} = -\left[ k_1(A_e+B_e)+k_{-1} \right]t.

Thus,

x=x0exp⁡{−[k1(Ae+Be)+k−1]t}.\boxed{ x = x_0 \exp \left\{ -\left[ k_1(A_e+B_e)+k_{-1} \right]t \right\}. }

The corresponding relaxation time is

1τ=k1(Ae+Be)+k−1.\boxed{ \frac{1}{\tau} = k_1(A_e+B_e)+k_{-1}. }

This result shows why relaxation experiments are useful for reversible reactions. A complicated nonlinear kinetic equation may become approximately first order when the system is only slightly displaced from equilibrium.

Temperature Dependence of the Equilibrium Constant

Section titled “Temperature Dependence of the Equilibrium Constant”

For an elementary reversible reaction,

K=k1k−1.K=\frac{k_1}{k_{-1}}.

If both rate constants follow Arrhenius behavior,

k1=A1exp⁡(−Ea,1RT)k_1 = A_1 \exp \left( -\frac{E_{a,1}}{RT} \right)

and

k−1=A−1exp⁡(−Ea,−1RT),k_{-1} = A_{-1} \exp \left( -\frac{E_{a,-1}}{RT} \right),

then

K=A1A−1exp⁡[−Ea,1−Ea,−1RT].K = \frac{A_1}{A_{-1}} \exp \left[ -\frac{E_{a,1}-E_{a,-1}}{RT} \right].

Taking logarithms,

ln⁡K=ln⁡A1A−1−Ea,1−Ea,−1RT.\ln K = \ln\frac{A_1}{A_{-1}} - \frac{E_{a,1}-E_{a,-1}}{RT}.

For a simple reaction coordinate,

Ea,1−Ea,−1≈ΔH.E_{a,1}-E_{a,-1} \approx \Delta H.

The temperature dependence of the equilibrium constant is therefore connected to the difference between the forward and reverse activation energies.

Kinetics determines how rapidly equilibrium is reached, while thermodynamics determines the equilibrium composition. For an elementary reversible reaction, the two descriptions are connected through the forward and reverse rate constants.

References

Section titled “References”
  • Atkins, P.; de Paula, J.; Keeler, J. Atkins’ Physical Chemistry.
  • Laidler, K. J. Chemical Kinetics.
  • IUPAC Compendium of Chemical Terminology (Gold Book), entries on equilibrium and relaxation methods.