Which R do I use? It’s a common question for chemistry students, and when it comes to the AP chemistry exam there is a choice of values for the ‘Universal’ Gas Constant. Currently appearing on an AP formula sheet near you, one can find the following;
a. R = 8.314 J mol-1 K-1
b. R = 0.08206 L atm mol-1 K-1
c. R = 62.36 L torr mol-1 K-1
So which value do we use in which situation?
There are a couple of ways to tackle this. Firstly a student could simply learn which value to use in which situation, by simply memorizing the correct pairings. For example, when using P V = n R T where the pressure is given in atm and the volume in L, then use b. Of course, this memorization method is not necessarily the preferable way to make sure that you are using the right R, since it does not involve any understanding as such, but having said that, if the choice is ‘learn the pairings parrot fashion with no understanding’, or ‘get the question wrong’, then I choose the former over the latter every time!
A ‘better’ way to handle the choice of R is to think about the units in each scenario. Below are a few common situations that should help students to clarify which R is in play.
1. Applying P V = n R T with ideal gases. As long as the pressure is in atmospheres and the volume in L, then you use b. since it is the only one with L and atm in its units. If the pressure is in torr or mmHg and the volume in L, then you use c . If you’d like to see a few more versions of R for different units of pressure, click here.
2. Applying ∆G° = – RT ln K when calculating Gibbs Free Energy. Not too difficult if you associate Gibbs free Energy with energy(!), and that therefore we must be dealing with Joules. Use a.
3. Although quantitative applications of the Arrhenius equation (e.g., ln k = -EA/RT + ln A) are now out, I suppose it is possible that one could be asked to understand that given the equation, -EA/R is the slope of the graph that plots various values of ln k on the y axis, against the corresponding values of 1/T on the x axis, and that multiplying the slope by -R can yield a value for activation energy, EA. Like #2, the simple link here is that we need J in the ultimate answer, so we need to use the R with joules in it. Use a.
It helps to know that the ln of a number is unitless (or at least to assume that and if you want to know more see this or speak to a mathematician), so the activation energy will pop out in J mol-1.
There’s one final twist that’s worth making a note of, that comes up most commonly in ∆G° equation and its use. ∆G° values are most often reported in terms of kJ, and when we use R = 8.314 J mol-1 K-1 in conjunction with kJ, there is the need to divide the R value by 1000 to convert J to kJ. Take a look at 2002B, 3d(i).
NOTE: This post is a re-write of one that was originally written in March of 2013 for the old exam. It has been updated to reflect the changes in the new curriculum, examined from May 2014 forward.
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