Not registered? Sign up. Publications Pages Publications Pages. Recently viewed 0 Save Search. Users without a subscription are not able to see the full content. Modern Thermodynamics for Chemists and Biochemists. Find in Worldcat. Go to page:. Table Most reactions that we encounter have equilibrium constants substantially greater or less than 1, with the equilibrium strongly favoring either products or reactants.
In many cases, we will encounter reactions that are strongly favored by thermodynamics but do not occur at a measurable rate. In contrast, we may encounter reactions that are not thermodynamically favored under standard conditions but nonetheless do occur under certain nonstandard conditions. In such cases, mixing the reactants results in only a physical mixture, not a chemical reaction.
In comparison, the analogous reaction of SiCl 4 with water to give SiO 2 and HCl, which has a similarly large equilibrium constant, occurs almost explosively. Although the two reactions have comparable thermodynamics, they have very different kinetics!
For example, consider the reaction of lead sulfide with hydrogen peroxide. One possible reaction is as follows:. Yet when lead sulfide is mixed with hydrogen peroxide, the ensuing vigorous reaction does not produce PbO 2 and SO 2. Instead, the reaction that actually occurs is as follows:. Thermodynamically, such reactions do not occur spontaneously under standard conditions. Nonetheless, these reactions can be made to occur under nonstandard conditions.
The real situation can be represented as follows:. This is why it is called the "steady state approximation. So the math for this scenario is as follows:. Pretty complicated rate expression, eh?! Depending on what tricks you use the steady state approximation is just one of them you can get some very crazy expressions.
If you understood the preceding example, you already understand all of the important ideas behind Michaelis-Menton kinetics. Michaelis-Menton does the same steady-state approximation math for a biological enzyme-substrate system. Enzymes are special proteins that catalyze biological reactions. Many enzymes break down food molecules into material from which your body can get energy. The enzyme called E has a little niche into which the substrate food molecule called S fits just perfectly.
We can represent the system as follows:. To solve this system, use the fact that the second step is the slow step to invoke the steady-state approximation. Note that there is no k -2 for this reaction: the enzyme will not catalyze the conversion of the product back to the substrate. Given all of these constraints, the math looks like what follows.
To make sure that you understand it, try to reproduce the answer from the given reaction sequence and the constraints detailed in this paragraph. If [S] is really big i. The last topic to consider before we leave kinetics and go back for a last look at thermo is integrated rate laws. Given the elementary steps, it is possible to integrate the corresponding rate law, using calculus to solve for the concentration of some reaction species as a function of time.
Test yourself heavily on both first order and second order rate law integration. From the equation for the elementary step, you should be able to figure out the concentration of the species as a function of time. Remember, this is just math. The chemical parts are only the first line and the last line of each derivation. Here are the answers you should be able to derive:. Actively test yourself on this!! You may be able to follow all the math, but could you reproduce it?
Common examples of first order reactions are radioactive decay processes. As must be the case for first order reactions, the rate of decay depends on the amount of radioactive stuff that's around at any one time. This results in an exponential decrease in the decay rate with time. When less stuff is around, it does not decay anywhere nearly as fast. When chemists talk about radioactive decay, they typically like to talk about half-lives. The half life of a substance is how long it takes for half of whatever's there to decay.
K Little k Thermodynamic, not kinetic Kinetic, not thermodynamic K doesn't really have units, though we often treat it as if it does. Rate constant changes with T and with catalyst. A catalyst would change the activation energy for the rate-determining step.
K is independent of the reaction mechanism. It is written from the products and reactants in the final, overall reaction equation. You can't know the rate law until you know the reaction mechanism and have identified the slowest step the bottleneck. Only reactants appear in the rate law.
Don't include solids or pure liquids in K because their essentially-constant concentrations are already absorbed into the equilibrium constant. Use the steady-state approximation.
The value of K depends on the stoichiometric coefficients of the equation to which it is referenced. Equilibrium may take a long time to be achieved! In order for this reaction to reach equilibrium, the concentration of B will have to increase beyond that of A up to the point where the rates of the forward and reverse reactions become equal. We can conclude that the net change in chemical potential energy is related to the equilibrium constant K eq for the reaction, whereas the magnitude of the activation energy must be related to the rate constants for the reaction.
Then we can readily derive the relationship between rate constants and the equilibrium constant for our simple first-order case. The rate laws for the forward and reverse reactions in this case are. At equilibrium, the forward reaction rate is equal to the reverse reaction rate. Setting these expressions equal when equilibrium concentrations are reached, we have. The right side expression is of course the equilibrium constant for this reaction, so we have shown in this case that.
Page updated Further consideration of the connection between the study of reaction rates chemical kinetics and equilibrium.
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