How to tell if reaction favors products or reactants

Equilibrium constants can be used to predict whether a reaction will favor the products or the reactants. K is the equilibrium ratio of products to reactants. The value for K is large (greater than one) when the products dominate the mixture and small (less than one) when the reactants dominate the mixture:

• K > 1: products are favored
• K = 1: neither reactants nor products are favored
• K < 1: reactants are favored

If the reaction favors the products, it will occur in the forward (left-to-right) direction. If K is very large, the reaction will occur mostly to completion, using up almost all the reactants. If the reaction favors the reactants, it will occur in the reverse (right-to-left) direction. If K is very small, the reaction will use up almost all the products and make them into reactants.

Reaction Quotients and Predicting Direction

The equilibrium constant is only used when a reaction has reached equilibrium. If a reaction is not at equilibrium, you can use the reaction quotient, Q, to see where the reaction is in the pathway:

• Q > K: reactants are favored
• Q = K: reaction is at equilibrium
• Q < K: products are favored

If you know the equilibrium constant for a reaction, and you know all the concentrations, you can predict in what direction the reaction will proceed.

Example 1: If 1.0 x 10-2 moles of hydrogen gas and 1.0 x 10-2 moles of iodine gas are placed in a 1-liter flask at 448 °C with 2.0 x 10-3 moles of HI , will more HI be produced?

Space-filling model of hydrogen iodide

Solution: The balanced equation for the reaction is:

$H_2 (g) + I_2 (g) \rightleftharpoons 2HI (g)$

The reaction quotient under these conditions is:

$Q = \frac{[HI]^{2}}{[H_{2}][I_{2}]}$

$Q = \frac {2.0 \cdot 10^{-3}}{(1.0 \cdot 10^{-3})^{2}}=0.040$

This value is smaller than the equilibrium value of 50.53, which tells us there is an excess amount of reactants present. Therefore, the reaction will proceed in the forward direction.

Example 2: If only 1.0 x 10-3 moles of H2 and 1.0 x 10-3 moles I2 had been used, together with 2.0 x 10-2 mole of HI, would more HI have been produced spontaneously?

Solution: You can verify that Q = 400. This value is greater than the equilibrium value: there is now too much of the products for equilibrium to exist. Therefore, the reverse reaction is favored.

Fundamentally, though it may seem intuitive, adding concentrations on either side of the reaction and comparing them to see which is larger isn't a measure of favourability, since $K_c$ is defined by the multiplication of the concentrations. It's like taking two rectangles and trying to measure which has a larger area by comparing their perimeters; it doesn't work in general unless you specify some additional constraint(s).

Under your definition of favourability by addition, you're weighing reagents and products unfairly, in some sense, because in the expression of $K_c$ there are two reagents and one product. This is a bit (if indirectly) reminiscent of trying to compare salt solubilities by comparing their $K_\mathrm{sp}$ directly instead of actually calculating their solubilities; it doesn't always work, because $K_\mathrm{sp}$ of salts are weighted by how many ions are released when dissolved.

It is not strictly true that the products would be favoured by your measure for any $K_c > 1$ even if the number of reagents and products were equal - just add one reagent in heavy excess. For example, consider the hypothetical reaction:

$$\ce{A + B -> C + D} \qquad \qquad K_c=\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}=1000$$

One possible solution for equilibrium concentrations is $[\ce{A}]=1\ \mathrm{M}$, $[\ce{B}]=10^{-5}\ \mathrm{M}$ and $[\ce{C}]=[\ce{D}]=10^{-1}\ \mathrm{M}$, and here $[\ce{A}]+[\ce{B}]>[\ce{C}]+[\ce{D}]$.

Postby Diana_Visco_1l » Thu Nov 24, 2016 11:54 am

When Kc>1 you can also say that kforward>kreverse and inversely if Kc<1 then k reverse>k forward. To favor either the reactants or the products in equilibrium is to say the formation of either the reactants or products is favored, as indicated by the rate constants. If a reaction is not at equilibrium, you can use the reaction quotient, Q, to see where the reaction is in the pathway. If Q > K, the reactants are favored. If Q < K, the products are favored. If Q = K, the reaction is at equilibrium. It's also important to realize that molecules never stop moving, even when equilibrium is reached. The larger the value for the equilibrium constant the more the reaction goes to completion. Irreversible reactions can be thought to have an infinite equilibrium constant so there are no reactants left.

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A common exam question that comes up in both Orgo 1 and Orgo 2 requires students to determine a reaction’s equilibrium position. The question usually looks something like this: 

Are the reactants (Keq < 1), or are the products (Keq > 1) favored in each of the below reactions:

Solving this type of question is easy with a little help from the Protocol and one more idea.

Applying the Protocol

We all know that nature hates energy–the lower the energy, the better. With this idea in mind, don’t forget we can use the Protocol to evaluate the relative stability of two negative charges, where the more stable negative charge is lower in energy.

Thus, one way to determine whether the reactants or products are favored in an equilibrium is to compare the stabilities of two negative charges on opposite sides of the equilibrium-arrows. Whichever side has the more stable negative charge is favored because this side is lower in energy.

One way to think about why this statement is true is to consider what happens once you get to the more stable negative charge. If you just got to a more stable position, you’re less likely to react in the backward direction because you’re happy where you are–a place of low energy. In other words: once you get there, you stay there!

In the above example, we’re putting the protocol to work by comparing a negative charge on an oxygen to a negative charge on a nitrogen. Since we’re comparing the charges on two atoms in the same row of the periodic table, we use electronegativity (#2 on the Protocol) to determine the relative stabilities.

A Notable Exception:

It’s worth noting that Orgo 1 introduces a major exception to the Protocol. Namely, Orgo 1 says you can use sodium amide (NaNH2) to deprotonate terminal alkynes. (Remember, this step is important for Orgo 1’s only path toward C-C bond formation.) I know, I know: too many words. Here’s a visual:

What you should notice is that the equilibrium favors the negative charge on the carbon instead of the nitrogen. This contradicts the Protocol’s prediction that a negative nitrogen beats a negative carbon. However, once you notice the hybridization of both atoms (pointed out in green, above) you can see that we have two competing stabilizing factors: the electronegativity of the atom and the electronegativity effect of the atoms’ hybridization.

This is an exception to memorize, and it’s easy to remember since we do it all the time to create C-C bonds.

Proton Transfer is #1

What if there aren’t two negative charges on opposite sides of the arrows? A second way to evaluate an equilibrium is to consider that the transfer of a proton from good bronsted acids to good bronsted bases is always favorable.

Conclusion

Once you understand the Protocol and the great favorability of proton-transfer, determining the equilibrium position of reactions is straightforward. If you come across equilibrium questions on your exam, don’t forget to ask yourself if you can compare two negative charges on opposite sides of the arrows. (And don’t forget to double-check you’re not looking at our notable exception.) 

If there aren’t two charges to compare, then consider that you may be looking at a proton-transfer between common orgo acids and bases (e.g. amines, strong acids, carboxylic acids).

Best of luck!