I don’t understand why the following question

The decomposition of nitrosyl bromide $(\ce{NOBr})$ proceeds by the following reaction:

$$\ce{2 NOBr(g) <=> 2 NO(g) + Br2(g)} \qquad K = 0.0142$$

Calculate the $[\ce{NOBr}],$ $[\ce{NO}],$ and $[\ce{Br2}]$ when $\pu{10.0 mol}$ of nitrosyl bromine is placed in a $\pu{5.00 L}$ closed vessel and allowed to decompose.

has the answers that it does:

$$ \begin{align} [\ce{NOBr}] &= \pu{1.585 M} \\ [\ce{NO}] &= \pu{0.415 M} \\ [\ce{Br2}] &= \pu{0.207 M} \end{align} $$

From what I understand

$$K = \frac{\prod a_\mathrm{products}}{\prod a_\mathrm{reactants}},$$

where $a$ is the activity of the products/reactants. The activity of a gas is approximately equal to the partial pressure of the gas divided by a reference pressure (usually 1 atm). This yields unitless quantities with the magnitude of the partial pressures of each gas. Thus

$$K = \frac{P_\ce{Br2}\cdot P_\ce{NO}^2}{P_\ce{NOBr}^2}$$

Converting from a partial pressure to a concentration can be done using the ideal gas equation

$$\frac{P}{RT} = \frac{n}{V} = M,$$

where $M$ is concentration. Thus,

$$K = \frac{[\ce{Br2}][\ce{NO}]^2\cdot RT}{[\ce{NOBr}]^2}$$

However, plugging the concentrations given into the solution of this problem doesn’t yield the given value of $K.$

Am I wrong in the way I am approaching this problem, or is the problem itself wrong? If the equilibrium constant was denoted $K_c$ instead of $K,$ the given solution would be correct, but I don’t think it’s correct to assume that $K_c$ is the same as $K.$

  • $\begingroup$ At a first glance I notice that the units don't seem to match - K should be dimensionless, but in your description of K by partial pressures it has a unit of pressure. So I think there is an error somewhere in your derivation. $\endgroup$
    – PoorYorick
    Jul 31, 2019 at 6:30
  • $\begingroup$ Does the problem not stipulate the value of T? $\endgroup$
    – Buck Thorn
    Jul 31, 2019 at 9:49
  • $\begingroup$ It's easier to think of $K$ as dimensionless, particularly if you plan to find the logarithm. $\endgroup$
    – Zhe
    Jul 31, 2019 at 13:03
  • 3
    $\begingroup$ The $K$ you are given is $K_c$ not $K_p$ as the values given do yield the equilibrium const. If you want to work out these values make up an ICE table with initial concentration $n/V =0.5$. The equilibrium constants are related as $K_C=K_P(RT)^{-\Delta n}$ where $\Delta n$ is the number of moles of products minus reactants according to stoichiometry , 1 in your case. The equilib. constants are always dimensionless as they are always divided by 1 M or 1 atm as appropriate. $\endgroup$
    – porphyrin
    Jul 31, 2019 at 16:24

3 Answers 3


The thermodynamic equilibrium constant $K$ (different from both $K_p$ and $K_c$) and the reaction quotient $Q$ are defined with reference to a standard state. When all reactants and products are at standard state, $Q = 1$. $K$ and $Q$ are related to the Gibbs energy of reaction via: $$\Delta_\mathrm{r}G = RT \ln\frac{Q}{K}$$ No matter which standard state you use, this relationship gives the same Gibbs energy of reaction (because $K$ and $Q$ are equally affected). However, depending on which standard state is used, $K$ and $Q$ will have different numeric values.

According to wikipedia,

In chemistry, the standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. In principle, the choice of standard state is arbitrary, although the International Union of Pure and Applied Chemistry (IUPAC) recommends a conventional set of standard states for general use.

The recommended standard state for a gas is, again according to wikipedia,

the hypothetical state it would have as a pure substance obeying the ideal gas equation at standard pressure ($\pu{e5 Pa}$, or 1 bar). No real gas has perfectly ideal behavior, but this definition of the standard state allows corrections for non-ideality to be made consistently for all the different gases.

Because neither the temperature nor the pressure is given in the problem the OP asked about, you can't use this standard state. Instead, you can use the standard state for a solute (still wikipedia):

For a substance in solution (solute), the standard state is the hypothetical state it would have at the standard state molality or amount concentration but exhibiting infinite-dilution behavior. The reason for this unusual definition is that the behavior of a solute at the limit of infinite dilution is described by equations which are very similar to the equations for ideal gases. Hence taking infinite-dilution behavior to be the standard state allows corrections for non-ideality to be made consistently for all the different solutes. Standard state molality is 1 mol kg−1, while standard state amount concentration is 1 mol dm−3.

What is the problem with this problem?

Am I wrong in the way I am approaching this problem, or is the problem itself wrong?

The problem is incomplete. If they use a funky standard state, they should say it explicitly.


Well maybe I'm an idiot, but I also get a different result.

First of all, the equilibrium constant is given by $$ K = \frac{\prod_p a_p^{m_p}}{\prod_r a_r^{n_r}} \,, $$ where $a$ is the activity of a reactant/product, $p$ is the index for products, $r$ is the index for the reactants, and $m_p$ and $n_r$ are the corresponding stoichiometric constants. So far, so good - you actually used this formula as well, as far as I can see, even though you didn't write it this way.

Now it's important to realize that the activity is dimensionless, and is given by $$ a_i = \gamma_i x_i = \gamma_i \frac{c_i}{c_\text{tot}} \,, $$ with the activity coefficient $\gamma$ ($=1$ for ideal gases) and the mole fraction $x_i$, which can be represented by the molar concentration $c_i$ of the reactant or product divided by the total molar concentration $c_\text{tot}$.

Using this and assuming $\gamma = 1$, we get $$ K = \frac{\prod_p c_p^{m_p}}{\prod_r c_r^{n_r}} \, \frac{\prod_r c_\text{tot}^{n_p}}{\prod_p c_\text{tot}^{m_r}}\,, $$ which in your case leads to $$ K = \frac{[\ce{Br2}][\ce{NO}]^2}{[\ce{NOBr}]^2} \, \frac{1}{[\ce{Br2}]+[\ce{NO}]+[\ce{NOBr}]} \,. $$ This is a dimensionless number, as desired. We know from your initial concentration $[\ce{NOBr}] = 2 \, \text{M}$ how many mol of N, O, and Br are involved. Solving this equation for the concentrations is a little bit annoying, but in the end I get for $K=0.0142$: $$ [\ce{NOBr}] = 1.48 \, \text{M} \\ [\ce{NO}] = 0.52 \, \text{M} \\ [\ce{Br2}] = 0.26 \, \text{M} \\ $$

I don't see where I could have gone wrong. I can only get the solution that you gave if I set the total concentration to $1 \, \text{M}$, and I don't see how that could be right.

Edit: After some digging, I found another definition of the activity which gives it as $$ a_i = \gamma_i \frac{c_i}{c^\ominus} \,, $$ where $c^\ominus$ is the concentration of the reactant or product in its standard state. This concentration is given as $c^\ominus = 1 \, \text{M}$, which then leads to the result given in the OP as the "correct" solution. However, I am not happy with this. It makes no sense and contradicts the other definition of the activity, in which the mole fraction is used. Since $c_i/c^\ominus$ is not a mole fraction, but some other random fraction, I don't think this is correct.

  • $\begingroup$ The activity depends on the choice of standard state. If the standard state is the pure substance, the mole fraction is a good approximation in some cases (when Roult's law is a good description). If the standard state is the pure gas at 1bar, you need to know the pressure. If the standard state is 1 M but exhibiting infinite-dilution behavior, the given data allows you to estimate the activity. $\endgroup$
    – Karsten
    Jul 31, 2019 at 17:11
  • $\begingroup$ @KarstenTheis, yeah. And I think the standard state that is used here is that the concentrations are related to $c^\ominus = 1\,\text{M}$, leading to the values given in the solution. But this is not satisfying to me, since I get different values with a different and, as far as I can see, equally valid approach. $\endgroup$
    – PoorYorick
    Jul 31, 2019 at 17:16
  • $\begingroup$ Without knowing the standard state, K is a meaningless number. $$$$ It is like saying the mass of a sample is 5. $\endgroup$
    – Karsten
    Jul 31, 2019 at 17:26
  • $\begingroup$ @KarstenTheis, I mean, you can do that. As long as you use natural units. $\endgroup$
    – PoorYorick
    Jul 31, 2019 at 17:41
  • $\begingroup$ ... and say so. Otherwise, someone could always go ahead and say "but I meant milligrams". $\endgroup$
    – Karsten
    Jul 31, 2019 at 17:44

$K_c$ is the equilibrium constant for reactions where reactants and products are all in the same phase, which is the case for your reaction. The subscript c implies that $K_c$ is expressed in terms of concentration, which has units of moles per volume. I'm not entirely sure on this, but I don't believe you can use partial pressures in the expression for $K$, because the total number of moles of gas changes during the reaction (2 moles of reactants become 3 moles of products), and the container has constant volume, so the partial pressures would change from the reaction.


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