{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture 26

# lecture 26 - Reversible first order reactions ...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Reversible first order reactions Although some processes can be approximately viewed as being irreversible, strictly speaking, there is no such thing as an irreversible reaction. If A can be converted to B then B can also be converted to A. Consider the reaction A = B In the absence of B we would have d[A]/dt =  ­k1[A]. In the absence of A B would be converting to A according to the law: d[B]/dt =  ­d[A]/dt = k ­1[B] If both A and B are present then: d[A]/dt =  ­k1[A] + k ­1[B] d[B]/dt = k1[A]  ­ k ­1[B] Introducing the extent of reaction, η=ξ/V, we write [A](t) = [A](0)  ­ η(t) [B](t) = [B](0) + η(t) and d[A]/dt =  ­dη/dt =  ­k1([A](0) ­η) + k ­1([B](0)+η) = η(k1+k ­1) + k ­1[B](0) ­k1[A](0) The solution of this differential equation (with the initial condition η(0)=0) is given by: η (t)= k1[ A](0) − k−1[ B](0) (1 − e − ( k1 + k−1 )t ) k1 + k−1 from which we can also find [A](t)= [A](0) ­ η(t) and [B](t) = [B](0)+ η(t) Of particular interest is the limit t → ∞ . In this case η (∞ )= and [A](∞)=([A](0)+[B](0))k ­1/(k ­1+k1) k1[ A](0) − k−1[ B ](0) k1 + k−1 [B](∞)=([A](0)+[B](0))k1/(k ­1+k1) If the system is allowed to evolve for a sufficiently long time, the concentrations of A and B achieve the above constant values, i.e., chemical equilibrium is attained. By dividing one equation by the other, we find that k1[A] (∞) = k ­1[B](∞) This equation could also be obtained by setting d[A]/dt and d[B]/dt to zero in the equations describing the time evolution of these two concentrations. In other words, the rate of the forward reaction (i.e., the one in the absence of products) should be equal to the rate of the reaction going in the opposite direction. This is a dynamic view of chemical equilibrium: both reactions, A→B and B→A are happening but their rates are equal so the concentrations of the reactants and the products do not change. The above equation for the equilibrium concentrations can be rewritten as [B]/[A] = k1/k ­1 On the other hand, [B]/[A] = K where K is the equilibrium constant. Thus we find k1/k ­1 = K This is a very important relationship. It tells us that the forward rate k1 and the backward rate k ­1 are not independent. Their ratio must be equal to the equilibrium constant K, which is determined by the thermodynamic properties of A and B. Imagine that we managed to speed up the conversion of A to B by using a catalyst. The above relation tells us that the conversion of B to A will also be faster in this case. Example: Suppose we have a container that has a protein solution and that this container is divided in two parts by a membrane that has asymmetric pores. Can we design a pore of such a shape that it would be more likely for the protein to cross it in one direction than the other (perhaps, a funnel shaped pore)? Then there would be a higher protein concentration on one side of the pore the on the other. Answer: No. the equilibrium constant for the “reaction”, in which proteins cross from one part to the other is K=1, regardless of the pore. ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online