Practice MCAT Questions:Â Equilibrium Constant K vs. RateÂ Constant k
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- Jun 02, 2016
Many of my students tend to get tripped up on the difference between the rate constant k in kinetics and the equilibrium constant K. They’re both big topics in the chemistry section of the MCAT, so it’s important to have mastered the difference between the two.
As such, I’ve created a few practice MCAT problems for you here, and will walk you through the solutions for each. Let’s do this.
Practice MCAT problems:
1. Given the following chemical reaction: N_{2(g)} + 3H_{2(g)} â†’ 2NH_{3(g) } and the information provided below:
a. Is the reaction at equilibrium?
b. If not, in which direction will the reaction move to reach equilibrium?
K_{eq} = 5.0
Substance | Current Partial Pressure (atm) |
N_{2} | 3.0 |
H_{2} | 2.0 |
NH_{3} | 5.0 |
2. Given the following chemical reaction: C_{3}H_{8(g)} + 5O_{2(g)} â†’ 3CO_{2(g)} + 4H_{2}O_{(g)} and the information provided below:
a. Determine the rate law equation for the reaction.
b. Calculate the value of k (with appropriate units) in the rate law equation
Experimental Data:
Rate (micromol CO_{2} produced) |
Starting Partial Pressures (atm) | |
C_{3}H_{8} | O_{2} | |
3.6 | 0.5 | 0.5 |
7.2 | 1.0 | 0.5 |
14.4 | 0.5 | 1.0 |
Discussion:
In order to answer this question, test takers need to be capable of distinguishing between two topics frequently tested by the MCAT: chemical equilibrium and chemical kinetics.
All too often, students mix aspects of these topics up, throwing away easy MCAT points. To be sure, there are a number of features of chemical equilibrium and kinetics that lead to this confusion. Below is a list of some of the biggest culprits:
- They both describe chemical reactions
- *On the MCAT, they often involve equations with reactants and/or products (in the case of chemical equilibrium or rate after the start of a reaction) included as parameters, raised to various exponents.*
- They are both called “k”, though one is capitalized.
In this post, we will review the fundamentals of chemical equilibrium and chemical kinetics as they pertain to the MCAT. We will then apply these basics in answering the question posed at the beginning of this post, clarifying the distinction between these concepts, as well as your approach to them. With the proper approach comes the correct answer on test day.
Chemical Equilibrium
This is a very important MCAT Chemistry topic.
At this state, the rate of forward and reverse chemical reactions are equal: reactant is being converted to product at the exact same rate that product is being converted to reactant. The concentrations of products and reactants are, therefore, unchanging. The reaction continues in both directions, but is effectively completed, as it’s no longer going anywhere.
For any given reaction, the point at which this equilibrium is reached can be described by a term known as Keq.
Keq = Equilibrium value/constant
For reaction aA + bB â†” cC + dD :
Reaction Quotient Q = ([C]^{c}[D]^{d}) / ([A]^{a}[B]^{b}) at any given point in the reaction
Equilibrium constant K = ([C]^{c}[D]^{d} / ([A]^{a}[B]^{b}) at equilibrium
This is where I want to really emphasize an important aspect of Keq.
When it comes to the Q or K values of equilibrium, the equation will always follow this exact form- the exponents will be the coefficients from the reaction formula. Products will always be in the numerator, reactants in the denominator. As we will see later, this is entirely different from kinetics, where the exponents are determined from experimental data.
Importantly, at equilibrium, K = Q.
What this means is that equilibrium is achieved for any set of reactant and product concentrations when the value for Q equals Keq.
Every chemical reaction has a specific Keq at a specific temperature. As is clear by the equation, there are an infinite number of reactant/product concentrations that can satisfy this equation. In other words, Q can be equal to K for an infinite number of combinations. But so long as this Q-K equality is satisfied, the reaction will be in equilibrium with forward and reverse reaction rates in exact opposition, and consequently with no changes occurring to any of the reactant or product concentrationsâ€¦. a state of dynamic equilibrium.
On the other hand, if the reactant and product concentrations generate a Q value that is not equal to K, then the current set of reactant and product concentrations is not in equilibrium and the reaction will continue to move in one of two directions, either forward, or reverse in order to proceed toward its final destination: equilibrium.
To determine which direction the reaction will move, one needs to compare the value of K and Q.
If Q>K, the numerator is too big; you need to move in a direction to reduce Q so that it will equal K. If a number is too big, you can lower it by decreasing the numerator, increasing the denominator or both — this corresponds to favoring the reverse reaction.
If Q<K, the opposite situation exists and the forward reaction is favored.
Chemical Kinetics
The term “k” will also come up in chemical kinetics, an important aspect of a chemical reaction that describes its rate, or the speed at which it occurs.
For the MCAT, you will often need to give an equation describing the rate of a reaction.
For every chemical reaction, the rate will be equal to some constant, k, multiplied by the reactant concentrations raised to corresponding exponents.
A couple of key points here: Whereas K in equilibrium is what the expression was set equal to, in kinetics k is a constant that appears on the “right side” of the equation, with the reactants raised to exponents. (i.e. for some reactions, A+Bâ†’C, Rate = K*(A^{x})*(B^{y}) where the values of K, X and Y are determined experimentally)
Secondly, in kinetics, only the reactants show up in the equation (when measuring initial rate, which will almost certainly be the case on the MCAT). The products will not. Also, the exponents are not given by the coefficient in front of the reactant. Instead they are determined experimentally. On the test you will deduce them from reaction rate tables, like the one above, that will be given to you.
In other words, if given a chemical equation: aA+bBâ†’cC the rate equation will take the following form:
Rate = K*(A^{x})*(B^{y}) where the values of K, X and Y are determined experimentally
Back to the example:
Here, I’ll methodically go through answering the questions.
1. Given the following chemical reaction: N_{2(g)} + 3H_{2(g)} â†’ 2NH_{3(g)}
a. Based on the information provided below, is the reaction at equilibrium?
b. If not, in which direction will the reaction move to reach equilibrium?
K_{eq} = 5.0
Substance | Current Partial Pressure (atm) |
N_{2} | 3.0 |
H_{2} | 2.0 |
NH_{3} | 5.0 |
This question is asking us to assess whether or not a reaction is at equilibrium, and to predict the direction the reaction will move in if it is not at equilibrium.
The information provided consists of a balanced chemical reaction, Keq for the reaction and a table of the current partial pressures. This should be setting off alarm bells that we need to compare Q with K.
For this reaction, all of the reactants/products are in the gaseous phase and as a result, all will be in the equation for Q. (Reactants/products in liquid or solid phases are left out while those in gaseous or aqueous phases are put in)
In this reaction, Q = ([NH_{3}]^{2})/ ([N_{2}]^{1}[H_{2}]^{3})
Plugging in the given values in our partial pressure table:
Q = ([5.0]^{2})/ ([1.5]^{1}[2.0]^{3})
Q = (25.0) / (1.5 * 8.0)
Q = 25.0 / 12.0
Q = 2.08
Thus, in this example Q < K (because Q = 2.08 and K = 5.0 as given)
As a result, this reaction is not at equilibrium. In order to reach equilibrium, this reaction must shift to the right, favoring formation of products. As the reaction shifts right, Q will increase until reaching 5.0 at which point the reaction has reached equilibrium.
2. Given the following chemical reaction: C_{3}H_{8(g)} + 5O_{2(g)} â†’ 3CO_{2(g)} + 4H_{2}O_{(g)}
a. Determine the rate law equation for the reaction using the information provided below.
b. Calculate k in the rate law equation.
Experimental Data:
Rate (micromol CO_{2} produced) |
Starting Partial Pressures (atm) | |
C_{3}H_{8} | O_{2} | |
3.6 | 0.5 | 0.5 |
7.2 | 1.0 | 0.5 |
14.4 | 0.5 | 1.0 |
This reaction is dealing with kinetics-reaction rates. The rate law will be determined experimentally using the table.
The first thing we need to determine is the order of the rate in terms of O_{2} and C_{3}H_{8}.
The easiest way to do this is to compare to columns in which only the reactant we are interested in changes.
C_{3}H_{8}: Comparing row 1 and 2, we see that C_{3}H_{8} changes, while O_{2} is unchanged. This makes these columns ideal for determining the order for C_{3}H_{8}. Specifically, we see that as C_{3}H_{8} is doubled, the reaction rate doubles. This proportional change tells us that the reaction is in fact first order with respect to C_{3}H_{8}.
O_{2}: Comparing rows 1 and 3, we see that as O_{2} is doubled from 0.5 to 1.0, the reaction rate quadruples to 14.4 from 3.6 (14.4/3.6 = 4.0). Thus, the rate is second order in terms of O_{2} because doubling its concentration results in a 2^2 or 4 fold increase in reaction rate.
*A useful check to make sure everything checks out is to compare columns where both reactants change and make sure the rate responds as predicted. We can do this by comparing columns 2 and 3. Between these columns, C_{3}H_{8} is cut in half. And O_{2} is doubled. The halving of C_{3}H_{8} should halve the reaction rate, the doubling of O_{2} should quadruple the reaction rate, and the combined effects should therefore be a doubling of the rate from column 2 to 3 (0.5 * 4.0 = 2.0). And as we see, the reaction rate goes from 7.2 to 14.4, a doubling. This should give us confidence that our calculated rate law coefficients are correct.
To calculate k, we simply need to set up our equation using the conditions from any of the columns given — K will be the same regardless of which column is chosen.
Rate = k * (C_{3}H_{8})^1 * (O_{2}) ^ 2
k = Rate / ( (C3H8)^1 * (O_{2}) ^ 2)
Plugging in from column 1,
k = 3.6 / ((Â½) * (Â½)^2)
k= 3.6 / (â…›)
k = 3.6 * 8
k= 28.8
The units for k can be solved using dimensional analysis:
Rate (Mol/S) = k* (Mol)^2 * (Mol)^2
Units of k = Mol^-3 * S^-1
Thus k = 28.8 M^03 * S^-1 and our overall rate reaction is:
Rate = 28.8M^-3 S^-1 (C3H8)^1 * (O2) ^ 2
To fully solidify your grasp of these concepts, work through more problems. And as you’re working through them, remain cognizant of which “k” you are dealing with. Keeping these concepts clear will result in a higher score on your MCAT. I hope this has helped! Good luck!