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At a certain temperature, $\ce{N2O5}$ dissociates as: \begin{align} \ce{N2O5 (g) &<=> N2O3 (g) + O2 (g)} & K_1 &= 4.5 \end{align} At the same time, $\ce{N2O3}$ also dissociates as: \begin{align} \ce{N2O3 (g) &<=> N2O (g) + O2 (g)} & K_2 &= \mathrm{?} \end{align}

When 4 moles of $\ce{N2O5}$ are heated in a flask of $\pu{1 L}$ volume, it has been observed that the concentration of $\ce{O2}$ is $\pu{4.5 M}$ at equilibrium. What are the equilibrium concentrations of the other products?

The combined reaction:

$$\ce{ N2O5(g) <=> N2O(g) + 2O2(g)}$$ $$K_\text{combined} = K_1 \times K_2 = 4.5 K_2$$

There are 4 moles of $\ce{N2O5}$ initially. If $x$ moles of $\ce{N2O5}$ dissociate to give $x$ moles of $\ce{N2O}$ and $2x$ moles of $\ce{O2}$, then plugging in the values in the $K_c$ expression for this reaction we get $$4.5 K_2 = \frac{4x^3}{4-x},$$ where $$x = \frac{4.5}{2 \pu{mol}}.$$

Solving for $K_2$, I get $K_2=\frac{81}{14}$, which is incorrect.

Is it allowed to first combine the reactions and then use the combined reaction?

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    $\begingroup$ I think there is sufficient information about where the OP's attempt to solve this went awry. It might help to see the "combined reaction", though. And maybe which species are in the flask after equilibrium is established. $\endgroup$ – Karsten Theis Mar 9 at 15:25
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    $\begingroup$ The misconception is the same as in this question. If there are two coupled equilibria, e.g. A reacts to B reacts to C, B might be non-zero, so you can't just combine the reactions to A reacts to C and neglect that there might be some B at equilibrium. I think this question merits an answer. $\endgroup$ – Karsten Theis Mar 15 at 15:48
  • $\begingroup$ Does this mean that to apply stoichiometry to an equilibrium problem it is necessary for the reactants to directly convert to products (elementary reaction)? $\endgroup$ – Sanom Dane Mar 15 at 15:59
  • $\begingroup$ You can, but you have to account for it. You can't say "x moles of N2O5 dissociate to give x moles of N2O", you have to say "x moles of N2O5 dissociate to give x moles of N2O3 or N2O", and use additional information to figure out how much is N2O3 and how much is N2O. $\endgroup$ – Karsten Theis Mar 15 at 16:46
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Different from Zhe's answer, I will use the approach of a combined ICE table using the model discussed in How can you use ICE tables to solve multiple coupled equilibria?. This results in a system with two unknowns that, in my opinion, gives a bit more insight into the problem as we are solving it. $x$ are changes due to dissociation of $\ce{N2O5}$ (first reaction), and $y$ are changes due to dissociation of $\ce{N2O3}$ (second reaction).

$$ \begin{array}{|c|c|c|c|c|} \hline &[\ce{N2O5}] & [\ce{N2O3}] & [\ce{N2O}]&[\ce{O2}] \\ \hline I & 4 & 0 & 0 & 0 \\ \hline C & -x & +x-y & +y & +x+y \\ \hline E & 4-x & +x-y & +y & 4.5 \\ \hline \end{array} $$

I have two unknowns, $x$ and $y$. Unless one of them is much bigger than the other (in which case I can first neglect the smaller one and come back to it later), I have to solve a system of two equations for $x$ and $y$ simultaneously. The first equation is already buried in the ICE table (subscript $eq$ is for equilibrium state):

$$ x + y = \ce{[O2]}_{eq} = 4.5 \ \ \ \ \text{or solved for y: }\ \ \ \ y = 4.5 - x$$

The second equation is via the equilibrium constant $K_1$:

$$ K_1 = 4.5 = \frac{[\ce{N2O3}]_{eq} [\ce{O2}]_{eq}}{[\ce{N2O5}]_{eq}} = \frac{[\ce{N2O3}]_{eq} \times 4.5}{[\ce{N2O5}]_{eq}}$$

Canceling 4.5 and rearranging gives $ [\ce{N2O5}]_{eq} = [\ce{N2O3}]_{eq}$, and substituting from the ICE table gives a second equation for x and y:

$$4-x = x - y\ \ \ \ \ \ \ \ \text{(or simplified } 2x = 4 + y)$$

Substituting the first equation into the second, we get:

$$ 2x = 4 + y = 4 + 4.5 - x$$ $$ x = 8.5/3 = 17/6$$

Substituting the value for $x$ back into the first equation:

$$ y = 4.5 - x = 4.5 - 17/6 = 10/6 = 5/3$$

Now we can tabulate the equilibrium concentrations and (to compare with Zhe's answer) the equilibrium constant $K_2$:

$$ c(\ce{N2O5})_{eq} = 7/6 M = 1.167 M$$ $$ c(\ce{N2O3})_{eq} = 7/6 M = 1.167 M$$ $$ c(\ce{N2O})_{eq} = 5/3 M = 1.667 M$$ $$ K_2 = \frac{[\ce{N2O}]_{eq} [\ce{O2}]_{eq}}{[\ce{N2O3}]_{eq}} = \frac{5/3 \times 4.5}{7/6} = \frac{45}{7}$$

We can check that the equilibrium concentrations add up to the initial concentration of 4 M, and that we obtain the correct value for $K_1$ when substituting our answers into the equilibrium expression.

Is it allowed to first combine the reactions and then use the combined reaction?

Yes, but we have to consider all species simultaneously because none of them are minor species. When calculating, for example, hydronium and hydroxide concentrations in a solution of a weak acid, we can set aside the autodissociation of water because hydroxide is a minor species, and adjusting the hydroxide concentration after calculating the weak acid equilibrium is fine because it does not change the hydronium concentration much, hydroxide being the minor species. Here, we can't do that and have to take the more comprehensive approach of considering the two equilibria simultaneously.

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  • $\begingroup$ One suggestion is actually to have 2 rows in the ICE table for the changes. This way, you can write separately the changes for each equilibrium to draw a better analogy to the single equilibrium problem. $\endgroup$ – Zhe Mar 18 at 12:23
  • $\begingroup$ @Zhe That is a great suggestion, but we need a better table type-setting tool on StackExchange so it's easier to make subdivisions in tables. $\endgroup$ – Karsten Theis Mar 18 at 13:05
  • $\begingroup$ Tables are kind of gross with every tool. :/ $\endgroup$ – Zhe Mar 18 at 14:01
  • $\begingroup$ @KarstenTheis@KarstenTheis:I take each reaction as independent equilibrium,I calculate the amount of $\ce{N2O3}~\text{and the amount of} ~\ce{O2}$ produced in the first reaction where the concentration of each equal$\pu{2.55234M}$,then I found the amount of the species in the second equilibrium where $[\ce{N2O}]=[\ce{O2}]=4.5-2.55234=\pu{1.94766M}$ and $[\ce{N2O3}]= 2.5534-1.94766=\pu{0.60468M}$ to calculate $K_2 = 6.3$. correct the solution. $\endgroup$ – Adnan AL-Amleh Mar 18 at 21:05
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Karsten's insight is correct. Your mistake is in assuming that you have either $\ce{N2O5}$ or $\ce{N2O}$ and $\ce{O2}$.

There are four concentrations of interest: $\ce{[N2O5]}$, $\ce{[N2O3]}$, $\ce{[N2O]}$, and $\ce{[O2]}$.

How can you relate these variables?

Well, you can apply conservation of matter in terms of oxygen and nitrogen.

For nitrogen:

$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$

For oxygen:

$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2 \ce{[O2]} = 5\times(4\ \mathrm{M})\tag{2}$$

Then, there are the equilibrium expressions:

$$K_{1}=4.5=\frac{\ce{[N2O3][O2]}}{\ce{[N2O5]}}\tag{3}$$

$$K_{2}=\frac{\ce{[N2O][O2]}}{\ce{[N2O3]}}\tag{4}$$

If you substitute $\ce{[O2]} = 4.5\ \mathrm{M}$ into (2), (3), and (4), you end up with:

$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$ $$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2\times(4.5\ \mathrm{M}) = 5\times(4\ \mathrm{M})\tag{2'}$$ $$4.5=\frac{\ce{[N2O3]\times(4.5\ \mathrm{M})}}{\ce{[N2O5]}}\tag{3'}$$ $$K_{2}=\frac{\ce{[N2O]\times(4.5\ \mathrm{M})}}{\ce{[N2O3]}}\tag{4'}$$

That's 4 equations and 4 unknowns, and it's all algebra from there.

Assuming I did that correctly, I got $$K_{2} = \frac{45}{7}$$

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