THE LAW OF CHEMICAL EQUILIBRIUM (MASS LAW)

The question arises: if you start with different total amounts of reagents (here 1-pentene and cyclopentane), what equilibrium states do you get? So, using SIR Q, try a few, using the same reaction at the same temperature. Here are two more examples.

Now you ask the class, "what do these equilibrium states have in common?"

Of course, here is where you see why 750° is a felicitous choice. It’s pretty obvious that in all the equilibrium states, the partial pressure of cyclopentane is exactly half that of
1-pentene. (You could use another temperature; the ratio of final pressures would still be constant, but it would not be so obvious.)

Here’s the appropriate moment for some conventions and definitions. The quantity [cyclopentane]/[1-pentene] is known as the reaction quotient of the reaction 1-pentene « cyclopentane. Its conventional symbol is Q. It can have any value whatever, even 0 (all 1-pentene) or ¥ (all cyclopentane), depending on what happens to be present at the moment.

At equilibrium, Q has an unique value, 0.50 in this example. This is the equilibrium constant, symbol K.

Since this reaction is reversible and can reach equilibrium, we usually write it with the double-arrow symbol instead of the single arrow.

As a reaction proceeds, Q changes; always in the direction of K. You may illustrate this, using the "show reaction quotient" option of SIR Q.

A number of interesting points may be raised here. How do you know if a reaction has reached equilibrium or is just running so slowly that you can’t see further change? What is the equilibrium constant at 750° of the reaction,
cyclopentane « 1-pentene? Think about it – we’ve actually run this reaction! (Some comments on conventions might be appropriate here.)

Now it’s time to proceed to more complicated reactions. You might choose a 1 = 1+1 or a 1 = 2 type (or an inverse) from the ones available. Here’s what you would get if you chose the 2-butene – ethylene reaction at 575° , and tried a selection of starting states.

The simple ratio [C₂H₄]/[C₄H₈] = 3 looks tempting, but the ratio is quite different in the other two cases.

So set the class to experimenting with various forms of Q for this reaction. It shouldn’t take them too long to discover that

the expression [C₂H₄]²/[C₄H₈] has the same value, 1.80 atm, in all three cases. So it appears that [C₂H₄]²/[C₄H₈] is the correct form of Q, and the equilibrium constant is 1.8 atm.

(It’s your call whether or not to assign units to Q. If you dig into a physical chemistry textbook, you’ll find that [A] is actually the numerical ratio of the activity of A to its activity in its standard state – so strictly speaking [A], Q and K are dimensionless pure numbers. Best to go along with your course textbook)

Having come up with a theory, that [C₂H₄]²/[C₄H₈] = 1.8 at equilibrium, you and the class chould test it by running the system to equilibrium from several new starting states.

Now it’s time to proceed to the reactions 2NO₂ « N₂O₄, C₃H₈ « C₂H₄ + CH₄ and CH₂O+ H₂ « CH₃OH. If people have trouble figuring out the form for the two reagent situation, you might point out that C₄H₈ « 2C₂H₄ may be written
C₄H₈ « C₂H ₄ + C₂H ₄.

Finally, you might proceed to the NH₃, NO and SO₃ reactions. All that’s needed is a straightforward application of principles already established.

At some point you’ll want to take a reaction and determine its equilibrium constant at a different temperature, to establish that the equilibrium constant is characteristic of,

• the reaction,
• the way it’s written,
• and the temperature.

 

Click to see how to elucidate electric polarity using SIR POLARITY

Click to see the initial study of the 1-pentene - cyclopentane reaction

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