# Chemical equilibrium textbook question

One textbook question that I came across is: How, the answer Doesn't make sense to me because when you add 3M of B, that's the initial position and not the equilibrium position. When you add 3 moles of B, the equilibrium shifts to the right by Le Chatelier's Principle and so the concentrations of of A decreases and C increases, which makes sense with the last statement "A possible new equilibrium condition... because k=2". Is the answer incorrect, or am I missing something here?

• How much A reacted, how much B, and how much C was formed? Jan 21 at 22:43
• The solution is humbug. It does not state any volume change and it is clearly not an equilibrium state. Jan 21 at 22:57
• Also proposing this solution means someone should not hire unqualified people for this kind of job (teaching), simply calculate the new euqilibrium condition given K = 2 from the given new condition to reach the desired values. Jan 21 at 23:03
• Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of at least essential parts. See also formatting math/chemistry expressions/equations. Jan 22 at 8:41
• As Poutnik suggests, please transcribe text within images into typed text. Jan 22 at 8:49

The volume is assumed constant (1 L). It is not specified whether the reaction occurs in the gas phase, that is largely irrelevant, and neither pressure nor temperature information is provided, that is also not relevant to the problem. The equilibrium constant is a concentration equilibrium constant.

The question does not require you to seek an algebraic solution for the computation of the equilibrium concentrations under the new conditions, after addition of more B. The objective is that you should use your understanding of chemical equilibrium and in particular stoichiometry, as well as some intuition (or rather an understanding of the mathematical relationship between concentration and amount of substance under constant volume) to assess whether a proposed new equilibrium in which $$\pu{[B]= 3 M}$$ can be achieved from the new starting conditions.

When volume is constant the concentration $$C_i$$ and amount of substance $$n_i$$ are strictly linearly proportional: $$C_i = \frac{n_i}V = \text{constant} \times n_i$$ In the present problem this proportionality is particularly simple because $$\pu{V= 1 L}$$ and it is possible to think in terms of concentration or amount of substance equally well if you choose units of mol and molar (remembering that the quantities employ different units).

In order to arrive at a concentration of $$\pu{[B]= 3 M}$$ from the starting concentration of $$\pu{[B]= 4 M}$$ you would have to reduce the concentration by $$\Delta \pu{[B]= -1 M}$$ which would simultaneously cause changes $$\pu{\Delta [A]= -1 M}$$ and $$\pu{\Delta [C]= +1 M}$$ resulting in final concentrations $$\pu{[A]= 1 M}$$ and $$\pu{ [C]= 5 M}$$. As this final concentration of C is inconsistent with an equilibrium point, we can logically conclude that $$\pu{[B]= 3 M}$$ cannot be achieved.

Unfortunately it is a somewhat poorly formulated question and largely unhelpful answer key.

### Statement 1.

Consider the reaction $$\ce{A(g) + B(g) <-> C(g)}$$ at equilibrium in a 1-L container with [A] = 2 M, [B] = 1 M, and [C] = 4 M.

This statement is neither true nor false, I guess we will assume it as a given. It's worth noting that the reaction changes the number of moles of gas in the container. This is a hint that changes in pressure will change the equilibrium of the reaction. It's also worth noting that we are given no information on whether $$\ce{A}$$, $$\ce{B}$$, $$\ce{C}$$, or mixtures thereof, are ideal gases. Since we aren't told to assume ideality, we won't. All we know is that at this particular equilibrium point, there is a total concentration of 7 M of gas.

### Statement 2.

To this 1-L container you add 3 moles of B.

This again is neither true or false. Or at least, if it's false, the rest of the question is meaningless so let's assume it's true.

### Statement 3.

A possible new equilibrium is [A] = 1, [B] = 6, and [C] = 3....

This is possible, but would require massive deviations from ideality of the three gases or their mixture. With 10 M of gas total in the container, the pressure will increase relative to the initial equilibrium with only 7 M total. For ideal gases, this should result in an increase in the concentration of C, since it lies on the side of the equation with the smaller number of total moles of gas (2 equivalents on the LHS and one on the RHS). The proposed new equilibrium has less C than the initial equilibrium. Thus it is not feasible, at least for ideal gases. But since we don't know if the gases are ideal, this is premature. We are left concluding that this equilibrium may be possible, but only with very, very, strong and unusual departures from ideal gas behavior.

Also, we aren't told anything about the enthalpy of the reaction or how the container is insulated. Is it held at constant temperature? Or is it impermeable to heat flow, i.e. the reaction occurs adiabatically? If the reaction releases or absorbs heat, when we perturb the equilibrium, the temperature will change.

### Statement 4.

... because in both cases $$K$$ = 2.

There is no reason to care particularly about the value of $$K$$ for this reaction, since the equilibrium depends on pressure and temperature, not just the concentrations of the reactants and product. Adding more moles of B will change the pressure and the temperature. $$K$$ is not a meaningful quantity for this reaction.

## Summary

The answer you were given makes very little sense.