However, sometimes our body isn't quite at an equilibrium. There might be too much glucose in our blood, or perhaps not enough. The reaction quotient is a handy way of looking at reversible reactions that haven't yet reached equilibrium.
- This article is about the reaction quotient, Q, in chemistry.
- We'll define the reaction quotient and look at its expression before seeing how it differs from the equilibrium constant, Keq.
- We'll then go through an example of calculating the reaction quotient.
- Finally, we'll take a deep dive into how the reaction quotient relates to Gibbs free energy.
What is the Reaction Quotient?
If you've read the articles "Dynamic Equilibrium" and "Reversible Reactions", you'll know that if you leave a reversible reaction in a closed system for enough time, it will eventually reach a point of dynamic equilibrium. At this point, the rate of the forward reaction equals the rate of the backward reaction and the relative amounts of products and reactants don't change. Provided you keep the temperature the same, the position of the equilibrium doesn't change either.
It doesn't matter whether you start with lots of the reactants or lots of the products - as long as the temperature remains constant, you'll always end up with fixed relative amounts of each. This is analogous to your body always trying to bring your blood sugar levels back to a fixed point.
We can express the ratio between the relative amounts of products and reactants using the equilibrium constant, Keq. Because the position of an equilibrium is always the same at a certain temperature, Keq is always the same too. At equilibrium, the value of Keq is constant.
However, reactions may take a while to get to equilibrium. What if we want to compare the relative amounts of reactants and products in a system that still isn't quite there? For this, we use the reaction quotient.
The reaction quotient is a value that tells us the relative amounts of products and reactants in a system at a particular moment, at any point in the reaction.
Types of Reaction Quotient
You should be familiar with the different types of Keq. They measure the amounts of substances in different systems of reversible reactions at equilibrium in different ways. For example, Kc measures the concentration of aqueous or gaseous species in an equilibrium, whilst Kp measures the partial pressure of gaseous species in an equilibrium. Likewise, we can also get different types of the reaction quotient. In this article, we'll focus on just two of them:
- Qc is similar to Kc. It measures the concentration of aqueous or gaseous speciesin a system at a particular moment.
- Qp is similar to Kp. It measures the partial pressure of gaseous speciesin a system at a particular moment.
For a reminder of Keq, check out "Equilibrium Constant". It's important that you understand the ideas within that article before you come to learn about Q.
Let's now move on to look at the expressions for Qc and Qp.
Reaction Quotient Expression
The expressions for the reaction quotients Qc and Qp are very similar to the respective expressions for Kc and Kp. But whilst Kc and Kp take measurements at equilibrium, Qc and Qp take measurements at any one time - not necessarily at equilibrium.
Qc Expression
Take the reaction \(aA + bB \rightleftharpoons cC + dD\). Here, the capital letters represent species whilst the lowercase letters represent their coefficients in the balanced chemical equation. For the above reaction, Qc looks a little something like this:
$$Q_C=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
Here's what that all means:
Square brackets show the concentration of a species at a given moment. Therefore, [A] means the concentration of species A.
The superscript lowercase letters are exponents, based on the coefficients of species in the balanced chemical equation. Therefore, [A]a means the concentration of species A, raised to the power of the number of moles of A in the balanced equation.
Overall, the numerator represents the concentrations of the products, raised to the power of their coefficients, and then multiplied together. The denominator represents the concentrations of the reactants, raised to the power of their coefficients, and then multiplied together. To find Qc, you simply divide the numerator by the denominator.
Notice how similar this expression is to the expression for Kc. The only difference is that Kc uses equilibrium concentrations, whilst Qc uses concentrations at any given moment:
$$K_c=\frac{[C]_{eq}^c[D]_{eq}^d}{[A]_{eq}^a[B]_{eq}^b}$$
$$Q_C=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
Qp Expression
Let's take the reaction again. But this time, instead of measuring concentration, let's measure the partial pressure of each species. This is the pressure it would exert on the system if it occupied the same volume on its own. To compare the ratio of partial pressures of gases in a system, we use Qp. Here's the expression:
$$Q_p=\frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$$
Let's break that down:
P represents the partial pressure of a species at a given moment. Therefore, (PA) means the partial pressure of species A.
The superscript lowercase letters are exponents, based on the coefficients of species in the balanced chemical equation. Therefore, (PA)a means the partial pressure of species A, raised to the power of the number of moles of A in the balanced equation.
Overall, the numerator represents the partial pressures of the products, raised to the power of their coefficients, and then multiplied together. The denominator represents the partial pressures of the reactants, raised to the power of their coefficients, and then multiplied together. To find Kp, you simply divide the numerator by the denominator.
Once again, notice how similar this is to the expression for Kp. The only difference is that Kp uses equilibrium partial pressures, whilst Qp uses partial pressures at any given moment:
$$K_p=\frac{(P_C)_{eq}^c(P_D)_{eq}^d}{(P_A)_{eq}^a(P_B)_{eq}^b}$$
$$Q_p=\frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$$
Like with the equilibrium constant, Qc ignores any pure solids or liquids in the system, whilst Qp ignores any species that aren't gaseous. It is simple, really - you leave them out of the equation altogether.
Reaction Quotient Units
Q takes the same units as Keq - which, as you might remember, doesn't have any units. Both Keq and Q are unitless.
Like Keq, Q is technically based on activities. A substance's concentration at any point in a reaction is actually its concentration activity, which is its concentration compared to the standard concentration of the species. Both values are typically measured in M (or mol dm-3), and this means that the units cancel out, leaving a unitless quantity. Partial pressure is similar - we actually measure pressure activity, which is the substance's partial pressure compared to a standard pressure. Once again, pressure activity has no units. Because both forms of Q are made up of unitless values, Q itself is also unitless.
Difference Between the Equilibrium Constant and the Reaction Quotient
Before we go any further, let's consolidate our learning by providing a summary of the differences between the equilibrium constant and the reaction quotient. We'll further break it down into Kc, Kp, Qc and Qp:
Fig.1-A table comparing the equilibrium constant and the reaction quotient
Reaction Quotient Example
Before we finish, let's have a go at calculating the reaction quotient for a particular reaction at a given moment. In the article "Using the Reaction Quotient", we'll then compare this to the reaction's equilibrium constant and see what it tells us about the reaction.
A mixture contains 0.5 M nitrogen, 1.0 M hydrogen and 1.2 M ammonia, all present as gases. Calculate Qc at this particular instant. The equation for the reversible reaction is given below:
$$N_{2\,(g)} + 3H_{2\,(g)} \rightleftharpoons 2NH_{3\,(g)}$$
Well, first we need to write an expression for Qc. As the numerator, we find the concentrations of the products, all raised to the power of their coefficient in the chemical equation and then multiplied together. Here, our only product is NH3, and we have two moles of it in the equation. Therefore, the numerator is [NH3]2.
As the denominator, we find the concentrations of the reactants, all raised to the power of their coefficient in the chemical equation and then multiplied together. Here, the reactants are N2 and H2. We have one mole of N2 and 3 moles of H2. Therefore, our denominator is [N2] [H2]3. Putting this all together, we find an expression for Qc:
$$Q_C=\frac{[NH_3]^2}{[N_2][H_2]^3}$$
Now, all we need to do is substitute in the concentrations given in the question, remembering that Qc has no units:
$$Q_C=\frac{[NH_3]^2}{[N_2][H_2]^3}$$
$$Q_C=\frac{[1.2]^2}{[0.5][1.0]^3}=2.88$$
Reaction Quotient and Gibbs Free Energy
In your studies, you might have come across Gibbs free energy. It is a measure of how thermodynamically favorable a reaction is, and relates to the reaction quotient Q with the following equation:
$$\Delta G=\Delta G^\circ +RTln(Q)$$
Note the following:
- ΔG is the change in Gibbs free energy, measured in J mol-1.
- ΔG° is the change in standard Gibbs free energy, measured in J mol-1.
- R is the gas constant, measured in J mol-1 K-1.
- T is the temperature, measured in K.
This can help you identify an equilibrium! If ΔG equals 0, then the reaction is at equilibrium.
That's the end of this article. By now you should understand what we mean by the reaction quotient and be able to explain the difference between the equilibrium constant and the reaction quotient. You should also be able to derive an expression for the reaction quotient based on a system of reversible reactions then use your expression to calculate the reaction quotient.
Reaction Quotient - Key takeaways
- The reaction quotient, Q, is a value that tells us the relative amounts of products and reactants in a system at a particular moment.
- Types of the reaction quotient include Qc and Qp:
- Qc measures aqueous or gaseous concentration at a particular moment.
- Qp measures gaseous partial pressure at a particular moment.
- For the reaction \(aA + bB \rightleftharpoons cC + dD\) $$Q_C=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
- For the same reaction, $$Q_p=\frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$$
- The reaction quotient is unitless.
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