# Equilibrium constant for isothermal double equilibrium “shift”

I do not understand how the solutions to the following high school chemistry question are valid:

A $$\pu{1 L}$$ vessel containing $$\pu{0.300 mol}$$ of dinitrogen tetraoxide gas is allowed to come to equilibrium. $$\pu{0.200 mol}$$ of nitrogen dioxide then added and the system allowed to reach equilibrium again. The final concentration of nitrogen dioxide in the vessel is $$\pu{0.600 mol L^-1}.$$ The temperature is constant throughout these experiments.

Calculate the equilibrium constant $$K_c$$ for this reaction.

### My approach

1. Draw up an ICE table where we account for the first equilibrium set up:

$$\begin{array}{lccc} \ce{&N2O4(g) &<=> & 2NO2(g)} \\ \text{I} & 0.3 && 0 \\ \text{C} & -x && +2x \\ \text{E} & 0.3-x && 2x \\ \end{array}$$

1. Draw a second ICE table to account for the second shift, resulting in the difficult to solve equation:

$$\begin{array}{lccc} \ce{&N2O4(g) &<=> & 2NO2(g)} \\ \text{I} & 0.3-x && 2x+0.2 \\ \text{C} & +y && -2y \\ \text{E} & 0.3 - x + y && 0.6 \\ \end{array}$$

### Provided solution

$$\begin{array}{lccc} \ce{&N2O4(g) &<=> & 2NO2(g)} \\ \text{I} & 0.3 && 0.2\\ \text{C} & && \\ \text{E} & && 0.6\\ \end{array}$$

How can they just put the $$0.2$$ as an initial concentration? It was not there initially. Conceptually, how is this equivalent to what was done in the question?

How can I be sure/convinced that adding the $$\pu{0.2 mol}$$ later in the process is actually going to produce the same final equilibrium as adding it to an intermediate equilibrium?

Buck Thorn's answer already addresses the conceptual ideas, and Sam202's how to solve it brute force. In this answer, I will first show an alternate way of approaching this using the conservation of mass. Second, I will show an easier way of filling the ICE table, and use the ICE table to illustrate why a set of related starting conditions (the dollars/pounds analogy in Buck Thorn's answer) lead to the same equilibrium.

### Conservation of mass

We calculate the amount of nitrogen atoms in the system. We start out with 0.6 mole (from 0.3 mole of dinitrogen tetroxide) and add another 0.2 mole (from 0.2 mole of nitrogen dioxide) for a total of 0.8 mole. If at equilibrium, 0.6 mole are in the form of nitrogen dioxide, the remaining 0.2 mole nitrogen atoms have to be in the form of dinitrogen tetroxide, showing that there are 0.1 mole of dinitrogen tetroxide at equilibrium.

### ICE table

First, the ICE table correctly given by the OP:

$$\begin{array}{lccc} \ce{&N2O4(g) &<=> & 2NO2(g)} \\ \text{I} & 0.3-x && 2x+0.2\\ \text{C} & +y && -2y \\ \text{E} & 0.3 - x + y && 0.6\\ \end{array}$$ And this equation is super hard to solve.

We can replace the $$-2y$$ with an expression containing $$x$$ because the value in rows "I" and "C" have to add up to the value in row "E".

$$\begin{array}{lccc} \ce{&N2O4(g) &<=> & 2NO2(g)} \\ \text{I} & 0.3-x && 2x+0.2\\ \text{C} & +y && 0.4 - 2x \\ \text{E} & 0.3 - x + y && 0.6\\ \end{array}$$

Then, we can replace $$+y$$ (via the stoichiometry) and $$0.3 - x + y$$ (via addition in the column). Or you solve $$-2y = 0.4 - 2x$$ for $$y$$, and substitute into the expressions still containing $$y$$:

$$\begin{array}{lccc} \ce{&N2O4(g) &<=> & 2NO2(g)} \\ \text{I} & 0.3-x && 2x+0.2\\ \text{C} & x - 0.2 && 0.4 - 2x \\ \text{E} & 0.1 && 0.6\\ \end{array}$$

As you can see, the $$x$$ is still in the table, but canceled out for the equilibrium concentrations, so we don't need to know it to calculate the equilibrium constant.

[from OP's comments to Buck Thorn's answer] How can I be sure that adding the 0.2 moles of NO2 later is going to produce the exact same result as adding it in initially?

There is an entire set of starting conditions that gives the same equilibrium. In equilibrium problems, there is no "order of addition" effect. You can add one substance first, then the other, or back and forth. As for the ICE table, I can take the equilibrium concentrations and show you the entire set of starting conditions that yield the same equilibrium (this ICE table is unusual in that we are calculating initial conditions from known equilibrium conditions):

$$\begin{array}{lccc} \ce{&N2O4(g) &<=> & 2NO2(g)} \\ \text{I} & 0.1 + z && 0.6 - 2 z\\ \text{C} & -z && 2z \\ \text{E} & 0.1 && 0.6\\ \end{array}$$

As long as I don't get negative values for the initial concentrations, I get the same equilibrium. So $$z$$ can go from 0.3 to -0.1. Here are a couple of examples:

$$\begin{array}{cccc} z &\ce{N2O4(g) &<=> & 2NO2(g)} \\ 0.3 & 0.4 && 0\\ 0.2 & 0.3 && 0.2 \\ 0.1 & 0.2 && 0.4\\ 0 & 0.1 && 0.6\\ -0.1 & 0 && 0.8\\ \end{array}$$

I can take the initial conditions in any row of the table, and will get the same equilibrium (one is already at equilibrium, with $$z = 0$$).

I should say that the initial conditions are a thought experiment anyway because as soon as you add all the reactants or all of the products, the reaction starts, so the "initial" concentrations are never present in reality.

• oohhh I think I get it now. No matter what I put in initially, I am still going to get an equilibrium of NO2 and N2O4. And thus the equilibrium constant must be the same - no matter what I start off with. Frankly, it wouldn't have mattered if I put 2000 moles of NO2 into it. The Keq will always be equal Oct 14, 2022 at 8:38
• One more question then. I agree the Keq for the 2 cases is the same. But, how do I know that the actual concentration of NO2 is the same using my method and the answer's? How can I be certain that it's not actually a different set of final concentrations that still makes the Keq fraction work? Oct 14, 2022 at 8:57
• @JohnHon Equilibrium is when $Q = K$. As the reaction goes in the forward direction, $Q$ will steadily increase (monotonically), and decrease for the reverse reaction. There is only one set of concentrations (for a given total mass of reactants and products) where $Q$ is equal to $K$, so there is a single solution. If you start with a different total mass, you will also reach equilibrium, but with different concentrations of species (and still, $Q = K$).
– Karsten
Oct 14, 2022 at 10:56

The order in which you add the reagents doesn't matter. The amount of dinitrogen tetraoxide reacted (x) is given by $$\pu{0.600 mol}=\pu{0.200 mol}+2x$$. The remaining amount is $$\pu{0.300 mol}−x$$. From these you can compute the equilibrium constant ($$x=\pu{0.200 mol}$$....).

Think of an analogy: I give you x dollars and y pounds. Whether you convert some of the dollars into pounds before or after I give you the pounds doesn't matter at the end (after the final equilibrium is established). You know how many pounds you had in the beginning and in the end, therefore you know the total dollars that were converted.

You tried to find an intermediate equilibrium, but lacked the information to do so (until computing the equilibrium constant from the final concentrations), and did not need to.

• Your analogy is interesting. Yes I agree that I can figure out how many dollars/pounds were converted given the initial and final. But how can we be sure it is the same if you give me the pounds later? How can I be sure that adding the 0.2 moles of NO2 later is going to produce the exact same result as adding it in initially? Oct 12, 2022 at 13:34
• Yes, the analogy works if you assume that you can predict where the equilibrium will lie in the end (you convert just the right amount of dollars before I give you the pounds). The catch is that the people who wrote the problem tell you what the final equilibrium result is (almost, they tell you half). Oct 12, 2022 at 13:59
• How you get to the final equilibrium is not important, provided you perform appropriate book-keeping. In the intermediate step you can convert an amount of N2O4 into an amount of NO2 that will result in a final equilibrium when you then add the NO2, that is, if in the intermediate step you convert (0.6-0.2)/2 moles of N2O4 into NO2. In reality it doesn't matter when the conversion happens, provided that the end result is the same. Oct 12, 2022 at 14:00

There is a way to solve this problem with the two-step equilibrium system you proposed, although it's a bit more complicated than the solution you posted.

Let:

$$A$$ represent $$N_2O_4$$

$$C$$ represent $$NO_2$$

Then the reaction becomes:

$$\ce{A <=> 2C}$$

From your second ICE table, we can see that:

$$2x+0.2-2y=0.6$$

Setting $$y$$ in terms of $$x$$:

$$y=x-0.2$$

Since Temperature is constant, $$Kc$$ will remain constant, so we have:

$$Kc=\frac{C_{C2}^2}{C_{A2}}=\frac{C_{C1}^2}{C_{A1}}$$

Rearranging:

$$\left(\frac{C_{C2}}{C_{C1}}\right)^2=\frac{C_{A2}}{C_{A1}}$$

Now, we substitute all variables in terms of $$x$$ and $$y$$:

$$\left(\frac{0.6}{2x}\right)^2=\frac{0.3-x+y}{0.3-x}$$

Substituting the expression for $$y$$ we derived earlier:

$$\left(\frac{0.6}{2x}\right)^2=\frac{0.3-x+x-0.2}{0.3-x}$$

$$\left(\frac{0.6}{2x}\right)^2=\frac{0.1}{0.3-x}$$

We can solve this quadratic formula and keep the positive value:

$$x=0.2374$$

Solving for $$y$$:

$$y=0.0374$$

Substituting into $$Kc$$ on your second ICE table:

$$Kc=\frac{C_{C2}^2}{C_{A2}}=\frac{0.6^2}{0.3-x+y}=\frac{0.6^2}{0.3-0.2374+0.0374}=\frac{0.6^2}{0.1}=3.6$$

Substituting into $$Kc$$ on your first ICE table:

$$Kc=\frac{C_{C1}^2}{C_{A1}}=\frac{(2x)^2}{0.3-x}=\frac{[2(0.2374)]^2}{0.3-0.2374}=\frac{0.2254}{0.0626}=3.6$$

Either way, we get the same answer:

$$Kc=3.6$$