The Reaction Quotient - Online Tutor, Practice Problems & Exam Prep

The reaction quotient \( Q \) helps determine if a chemical reaction is at equilibrium. If \( Q = K \), the system is at equilibrium; if \( Q < K \), the reaction shifts forward, increasing products and decreasing reactants. Conversely, if \( Q > K \), the reaction shifts backward, increasing reactants and decreasing products. This balance is crucial for understanding dynamic equilibrium in chemical reactions, where the equilibrium constant \( K \) represents the ratio of products to reactants, excluding solids and liquids.

The Reaction Quotient

So, our reaction quotient q is just a tool that we use to see if our chemical reaction is at equilibrium. We're going to say here if the reaction quotient Q is equal to the equilibrium constant K, then our reaction is at equilibrium. Q is just like K, it equals products over reactants. And just like K, it ignores solids and liquids. Here in this example, we have A gas giving us B gas, so neither one is a solid or liquid. So the equilibrium expression or Q expression in this case would just be b/a. You'd get values for your q. You'd plug them in to find out what your q value is. If you figured out what your q value is and you saw that your equilibrium constant equal the same exact number, we would be at equilibrium. In this case, I've just come up with the value of being equal to 50. So again, when q equals k we are at equilibrium. Now when q does not equal k, we can use q to determine which direction our reaction will shift to get to equilibrium. Now we're going to say here the direction our reaction shifts determine whether our reactants or products are increasing or decreasing. Remember k, your equilibrium constant k is your whatever q is, q will always shift to get to k. Whatever q is, q will always shift to get to k. For example, let's say that we did q equals products over reactants and we determined q was equal to 10. Our k is still equal to the value of 50. So in this case, q is less than k. Wherever k is, q will shift there in order to reach equilibrium. So here, q would shift in the forward direction to reach k so that it can get to equilibrium. The direction it shifts on the number line is the same direction it shifts in the equation. So here we have A gas gives us B gas again. In the number line, we shifted in the forward direction to get to K, so we're going to shift in the forward direction in our reaction. Wherever we're shifting is increasing an amount. And if that side is increasing an amount, that means the other side has to be decreasing an amount. Remember, there's a balance involved They both can't be increasing or decreasing together. It's kind of like a balance. Now in the other example, here I've calculated q again. I did q equals products over reactants. I used given values and I figured out q was equal to 140. In this case, q is larger than k. But we still do the same thing. Q will always shift to get to k. It'll always move in the direction to get to k so they can reach equilibrium. Here on the number line, we shift in the reverse direction to get to k to get equilibrium. If I shift in that direction on the number line, I shift in the same direction in my equation. Again, wherever I'm shifting will always be increasing. So I'm shifting to the reactant side so the reactant side is increasing. Since the reactant side is increasing, that means the product side is decreasing. So remember, q is just a tool that we use to see if we're at equilibrium. If q equals k, we're at equilibrium, there will be no shifting. If q is less than or greater than k, then shift in some direction to get to equilibrium. Remember the fundamental steps we see here so that we can answer any question asking us what side is increasing or decreasing, which direction will our reaction shift to get to equilibrium. Now that we've established these fundamentals, we'll attempt to do the example question on the bottom. So click on to the next video and see how I approach this same exact question that we have here on the bottom.

The Reaction Quotient

Consider the hypothetical reaction below. So we have 3 moles of A gas reacts with 1 mole of B gas to produce 2 moles of C gas. Here we're told the equilibrium constant equals 1.5 times 10 to the negative 3. We're asked which one of the following statements is correct if the initial concentrations are A equals 0.85 molar, B equals 0.36 molar, and C equals 0.005 molar.

All right. They are giving us our k value. They are asking, are we at equilibrium? Is such and such increasing or decreasing? These are clues that you need to figure out what q is. Once we find q, we can compare it to k. Again, they'll give you your k value. They'll give you some other new numbers which will help you figure out what q is. Once you do that, you'll know which direction your reaction is shifting and therefore what side is increasing versus what side is decreasing.

So q is just like k. It equals products over reactants. We ignore solids and liquids. Here we have C squared divided by A cubed because remember the coefficients become powers times B. All we do now is we insert the values.

So C is 0.005 which will be squared divided by A, which is 0.85 cubed times 0.36. When we plug all that in, that gives us 1.13×10-41.13 \times 10^{-4}. Now that we have q, we can compare it to k.

Set up your number line. So k is where we want to be. It's to the negative 3. Q is to the negative 4, so q is smaller. So we'll have to shift in the 4 direction to get to k. If we shift in that direction on the number line, we shift in the same exact direction in our reaction. We're moving towards the product side so the product side will be increasing in amount and my reactant side will be decreasing in amount.

All right, so here at equilibrium, the amount of C will increase. Yes. That's true because we're moving towards products so products would increase. So A is true. At equilibrium, the amount of A will increase. No, we're shifting away from reactants so it should be decreasing. At equilibrium, the amount of B will increase and the amount of C will increase. We're moving away from B so it decreases plus both sides can't increase and both sides can't decrease. So that's not a possibility. Here the amount of A and B we said would both decrease. And then the reaction could only be at equilibrium if q equaled k. Here, k is 1.5×10-31.5 \times 10^{-3}. Q was definitely not that number. Since they did not equal each other, we are not at equilibrium.

Out of all the choices, only option A is the correct choice.

Reaction Quotient Calculations 1

Here we're told that the reaction of 2 moles of carbon dioxide decomposing to form 2 moles of carbon monoxide and 2 moles of oxygen gas has an equilibrium constant (K) of K=7.22∙10-4 at 400 Kelvin, while the reaction quotient (q) is q=6.63∙10-2. Initially, we have 0.20 atmospheres of CO₂, 0.30 atmospheres of CO, and 0.15 atmospheres of O₂.

Given that q vs. k indicates that q is larger (since -2 is a higher exponent than -4), the system will shift to decrease q to match k. In other words, the reaction will shift to increase the concentration of reactants (CO₂) and decrease the concentrations of products (CO and O₂). As a result, the partial pressure of CO₂ is expected to increase, making it greater than its initial value of 0.20 atmospheres, while the partial pressures of CO and O₂ should decrease from their initial values of 0.30 and 0.15 atmospheres respectively.

Let's review the statements:

• The pressure of CO₂ will be greater than 0.20 atmospheres: True (reactant increasing due to the shift).
• The pressure of CO will be less than 0.30 atmospheres: True (product decreasing).
• The pressure of O₂ will be less than 0.15 atmospheres: True (product decreasing).
• The reaction will favor reactants: True (as it shifts toward reactants).

All the statements are actually true. Note that if any statement suggested that the pressure of CO or O₂ would increase, or that the pressure of CO₂ would decrease, those would have been untrue. However, the provided analysis correctly matches the expected changes due to the reaction shift.

Reaction Quotient Calculations 1

For the following reaction, we have 3 moles of hydrogen gas reacting with 1 mole of nitrogen gas to produce 2 moles of ammonia gas as a product. Here we're told that the equilibrium constant for the reaction equals 25. Now it says at a particular time, the following concentrations are measured for the given compounds. Here H₂ becomes 0.005 molar, N₂ is 0.170 molar, and NH₃ is 3.12 × 10² molar. Which of the following statements is true? So here they're talking about things increasing or decreasing. The only way we can determine that is to first figure out which direction my reaction shifts. Does it move in the forward direction to favor products or does it move in the reverse direction to favor reactants?

To be able to determine which direction it shifts, we need to find out what q is and then compare it to k. So q is just like k. It equals products over reactants. So that's NH₃² divided by H₂³ × N₂. We're going to plug in these initial concentrations here to help us determine what q is. So that's (3.12 × 10²)² divided by (0.005)³ × 0.170. So when we plug all that in, that's going to give me as an answer, 4.58 × 10¹² for my q value. Now just compare q to k. So on my number line, k is where we want to be, so it's in the middle. So it's 25. Q is so much larger. It's 4.58 × 10¹². So remember, q always shifts to get to k. So it's going to shift this way to get to k which means in my reaction it shifts the same way. So this side here would be increasing over time and this side here would be decreasing over time.

Alright. So here, which of the following statements is true? Here the concentration of H₂ will increase. That is true because we're heading towards products. We know that's true. The equilibrium constant will increase. Your equilibrium constant is k. The only time that your equilibrium constant can be changed is if we're adjusting the temperature. Concentration of equilibrium constant will still be 25. The concentration of NH₃ will increase? No, we're shifting away from it so it will be decreasing. The concentration of N₂ will decrease. No, it should be increasing over time since we're shifting towards the reactants. No change will occur. This would only happen if we are at equilibrium. We are at equilibrium when q equals k. But q and k are definitely not the same value, so a change will occur. Out of all the choices here, we see that option A is the correct choice.

Remember, your reaction quotient is just used as a way of determining are you at equilibrium? When q equals k, we are at equilibrium and so your reaction will not shift in the forward or reverse direction. If q is different from k, we have to compare q to k and see which way does Q move to get to K. This determines if you'll move in the forward direction to favor the creation of additional product or reverse where you're going towards reactants to create more reactants. Just remember the fundamentals in terms of q and how to compare it to k and you'll be able to determine which side is increasing or decreasing within any given chemical reaction.