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[TOP]
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7. Empirical Reaction Kinetics (ii) (1-3) Reactions approaching equilibrium [ TOP ]
Reactions approaching equilibrium have to be treated in a special way to account for the fact that in proximity to equilibrium conditions, the reverse reaction becomes important. Let us consider a first order reaction close to equilibrium: The concentration of A is:
(1) Exactly the same conclusion can be reached (and far more quickly) by noting that at equilibrium, the forward and reverse rates must be equal, i.e.: (2) This equation is a particularly significant as it relates a thermodynamic property (the equilibrium constant) to quantities relating to rates. [ TOP ]
(1-4) Disturbing equilibria - Relaxation methods
The term relaxation denotes the return of a system to equilibrium, and in chemical kinetics, it is used to indicate the re-adjusting that a reaction has to make in response to a sudden (normally) disturbance of an equilibrium. Sudden changes could include:
Let us consider the scenario of a sudden temperature increase applied to the equilibrium A B (see Fig. 1). Let ka1 and kb1 be the rate constants at the initial temperatures T1, that is, prior to the temperature jump, the net rate if change of [A] is given by: which under equilibrium conditions: i.e.: Now let ka2 and kb2 be the rate constants at the final temperatures T2, that is, after to the temperature jump. As before, once we reach the new equilibrium at temperature T2 we have: Note that this state of equilibrium is reached at a rate that depends that depends solely on the new rate constants ka2 and kb2, that is from:
we obtain (for x in the region between 0 and '[A]eq1 - [A]eq2'): since: Also, since [A] = [A]eq2 + x, then: d[A] = dx, i.e.: Thus we have: i.e.: which becomes: i.e. by raising to the power of e: or: where t is known as the relaxation time given by: Note that through the relaxation time, t, and the new equilibrium constant we may derive ka2and kb2 by simultaneous equations ( K2 = ka2/kb2).Similar equations may also be derived for other types of elementary reactions. For example for A B + C where the forward is first order, and the reverse is second order overall, we have: i.e. at equilibrium 1: whilst in between equilibrium 1 and 2 we have:[A] = [A]eq2 + xi.e.: where: i.e.: i.e.: which since: we have: where is we assume that x2 is approximately zero, then: i.e.: i.e.: or: where:
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CH237 - Chemical Thermodynamics
and Kinetics
Dr.
Joseph N. Grima, Department
of Chemistry
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7. Empirical Reaction Kinetics (ii) (1-3) Reactions approaching equilibrium [ TOP ]
Reactions approaching equilibrium have to be treated in a special way to account for the fact that in proximity to equilibrium conditions, the reverse reaction becomes important. Let us consider a first order reaction close to equilibrium: The concentration of A is:
(1) Exactly the same conclusion can be reached (and far more quickly) by noting that at equilibrium, the forward and reverse rates must be equal, i.e.: (2) This equation is a particularly significant as it relates a thermodynamic property (the equilibrium constant) to quantities relating to rates. [ TOP ]
(1-4) Disturbing equilibria - Relaxation methods
The term relaxation denotes the return of a system to equilibrium, and in chemical kinetics, it is used to indicate the re-adjusting that a reaction has to make in response to a sudden (normally) disturbance of an equilibrium. Sudden changes could include:
Let us consider the scenario of a sudden temperature increase applied to the equilibrium A B (see Fig. 1). Let ka1 and kb1 be the rate constants at the initial temperatures T1, that is, prior to the temperature jump, the net rate if change of [A] is given by: which under equilibrium conditions: i.e.: Now let ka2 and kb2 be the rate constants at the final temperatures T2, that is, after to the temperature jump. As before, once we reach the new equilibrium at temperature T2 we have: Note that this state of equilibrium is reached at a rate that depends that depends solely on the new rate constants ka2 and kb2, that is from:
we obtain (for x in the region between 0 and '[A]eq1 - [A]eq2'): since: Also, since [A] = [A]eq2 + x, then: d[A] = dx, i.e.: Thus we have: i.e.: which becomes: i.e. by raising to the power of e: or: where t is known as the relaxation time given by: Note that through the relaxation time, t, and the new equilibrium constant we may derive ka2and kb2 by simultaneous equations ( K2 = ka2/kb2).Similar equations may also be derived for other types of elementary reactions. For example for A B + C where the forward is first order, and the reverse is second order overall, we have: i.e. at equilibrium 1: whilst in between equilibrium 1 and 2 we have:[A] = [A]eq2 + xi.e.: where: i.e.: i.e.: which since: we have: where is we assume that x2 is approximately zero, then: i.e.: i.e.: or: where:
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E-mail me at jgri1@um.edu.mt |