Back to Thermodynamics

Why is equilibrium such a central concept in chemical thermodynamics? Well, equilibrium describes what happens to reversible reactions, and reversible reactions are prevalent in God's world. We need useful tools to describe what we observe. Among other things, this helps us predict and even control outcomes. Molecules and particles are constantly moving at the temperatures of our experiments and life around us. Their collisions can lead to reaction. Once product is formed, it doesn't just sit there. Product substances collide, too, and can result in re-forming reactant. There they are -- reactants and products co-existing in a mixture, and still undergoing changes. This can keep going on... to a point. Once the rate at which reactant forms product matches the rate at which product changes back to reactant (rate slower than the initial activity of the reaction), the dynamic mixture achieves a balance point. Ongoing reactions, but no increased amounts of anything produced. The composition of the mixture is static. Good, then we can measure or calculate how much of any reactant or how much of any product is in the equilibrium mixture! K helps us. K is the ratio of products over reactants. Multiply ALL the product amounts and divide that by the multiple of ALL the reactant amounts. So, since products are in the numerator of this ratio, then a large K value (> 1) means there are more products in the mixture than reactants! It is useful to know the value of K for any equilibrium reaction. Ka is the value give to an acid in water equilibrium reaction. Different acids will have different Ka values. In class examples (and in lab), we will calculate K given the equilibrium amounts of reactants and products. We will also do the opposite -- calculate equilibrium amounts by using a known K value -- this type is more involved (and often needs the wonderful quadratic formula :-).

Answers for Pink Practice Sheet and MCAT questions

If you are working through the pink practice problems in preparation for the exam, here are some answers for you to compare... 2. Yes since k changes with temperature., 3. 250 kJ/mol, 4.a 103 kJ, c first, d 779 sec, e 2.36 x 10-3 M/s, 5.b 344 (L/mol)^3/2(1/s), c 24.0 M/s, 6.a X, b 40 kJ, c -30 kJ, d 70 kJ, e Y, f X, g Y, h both. For the heat of reaction practice problem, both methods happen to give the same answer. If you have worked on the MCAT questions on this material, here are some answers: Passage 16 A, D, C, D, A, B, C, Passage 18 C, A, D, C, C, D, Passage 24 D, A, D, A

Break for Kinetics

If changes in thermodynamic state functions do not depend on how reactants convert to products, then we can't use them to tell us about a reaction's in-between details. Studies in Kinetics (motions of molecules) can lead us toward this information. Empirical Kinetics is all about trying to come up with a rate expression (rate law) for a chemical reaction of interest. A rate expression (law) is a general summary of how dependent a reaction's rate is on the concentration of each reactant at a specified temperature. To come up with a rate expression, we need to determine the order with respect to each reactant, and then get a value for the rate constant (k) of the reaction at the given temp. How to do? 1. Use the method of initial rates: You did this in lab. Get initial rate vs. [reactant] data, or 2. Try to plot different [reactant] vs time plots to see if data matches up with one of the "integrated rate equation" cases Kayli outlined in the previous post for m = zero, first or second order. Then you can get a value for rate constant (k) from the slope of the straight-line plot at a given temperature. Sound easy enough? Let's try to avoid other cases :-)

Integrated Rate Equations

Hi everybody!

Today we learned about the different graphs for different orders of reactions. A zero order reaction rate has [A] on the y-axis and time on the x-axis. A first order reaction rate has ln[A] on the y-axis and time on the x-axis. Lastly, a second order reaction rate has 1/[A] on the y-axis and time on the x-axis.

A way that may help you remember that the first order has the ln[A] on the y-axis is that the 'l' in 'ln' looks like a 1! Hope that helps a little bit!

Hess's Law

A scientific law is a simplified, general summary of what we can readily observe in God's world -- His ordinary way of sustaining and governing the creation. Hess observed that changes in state properties like enthalpy can be determined either directly or by adding together a series of steps in an overall change in any order. In your experiment, you switched the order of the three reactants to test this observation. Was the overall deltaH per gram of NaOH about the same regardless of the order in which you mixed reactants? Hopefully! Hess's observation and the idea of state property are two somewhat redundant concepts. We can observe what Hess did BECAUSE of state properties. Any property that can be calculated simply as a change between final (products) and initial (reactants) states is a state property. H, T, P, V... are all state properties. q and w are path properties. Heat and Work are the pathways of energy transfer so we don't talk about delta q or delta w. Either energy is transferred in an organized way or a disorganized way. There is no meaning to a change in the transfer of energy -- transfer IS the change. Hess's Law allows us to cleverly calculate properties such as deltaH of a reaction (heat of reaction) using heats of formation values. It is "clever" since we do not have to carry out a direct reaction, instead we relate experimental values indirectly to get what we want. We end up with a more general and accurate method for heat of reaction than by worrying about gas-phase average Bond Enthalpies. We'll use Hess's Law as a tool to determine values for other state property changes like entropy (S) and Gibbs free energy (G) coming up...

Upcoming Quiz

Hey guys!
So we have a quiz coming up on Friday on Thermochemistry, and I am sure we are all nervous, myself included.
I would say that one of the best ways to study is to get together with a group of people and to talk about the concepts. Being able to think and fully explain a concept to someone else is one of the best ways to integrate the material into your mind.
Also, one of the best things for me, personally, is just to do practice problems.  Go beyond the homework. Practice the others from the book that are similar to those that you have problems on. It is a HUGE help.