Equilibrium constant for isothermal double equilibrium “shift”

Question

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:

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

2. 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?

Answer 1

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.

Answer 2

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.

Answer 3

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:

$$Kc=3.6$$