Calculating K For Overall Reaction: Study with Video Lessons, Practice Problems & Examples

Chemical reactions often occur in multiple steps, each with its own equilibrium constant. To calculate the equilibrium constant (K_eq) for the overall reaction, you must consider the equilibrium constants of the individual steps. The process involves locating each compound from the overall reaction within the partial reactions, ensuring they match in number and phase. If a compound appears in multiple reactions, skip it. You may need to reverse or multiply the partial reactions, affecting the equilibrium constant. After aligning the reactions correctly, combine them, canceling out intermediates—compounds that appear as both reactants and products. The overall Keq is then found by multiplying the constants of the partial reactions. This method ensures the accurate determination of the equilibrium constant for complex reactions, which is crucial for predicting the reaction's behavior and understanding chemical kinetics and thermodynamics.

Calculating Keq of the Overall Reaction

Now recall that many chemical reactions cannot be carried out in one step but instead require multiple steps to get the final product. Here we're going to say that K_eq is our equilibrium constant of the overall reaction and it is the product of the equilibrium constant values of each of these multiple steps. So if we take a look at this example question it says, calculate the equilibrium constant for the overall reaction. Here we have 1 mole of carbon solid reacting with 1 mole of carbon dioxide gas to produce 2 moles of carbon monoxide gas. Here we need to find the overall equilibrium constant when given the following partial reactions, each with their own equilibrium constant. Alright, so how do we approach something like this? All we have to do is follow the steps provided before us. So step 1 is we're going to start with the first compound in the overall reaction and locate it in the set of partial reactions. So our overall reaction, we have carbon solid here. It is a reactant, there's 1 mole of it. We need to find it within our partial reactions below. If we look, here it is right there. But there are a couple of things that are wrong here. First, well really the main thing that's wrong with it is that it's a product here. We need it to match the same position as our overall reaction. So compounds from partial reaction must match the number, so they have to have the same coefficients, and also they have to have the same phase as the one from the overall reaction. Right. So both of them are 1 mole, 1 mole, that's good and they're both the same phase, solid and solid. It's important to remember the phases need to match because sometimes you'll have water as a liquid in one partial reaction and water as a gas in another partial reaction. Alright, this may require you to reverse multiply or divide the partial pressures which will affect your equilibrium constant in blue. Okay. Now step 2, you keep moving on to the next compound in the overall reaction once you locate all compounds in the partial reactions. Skip compounds found in multiple reactions. Alright. So let's just do what it says there. We need to follow the overall reaction, so we need 1 mole of carbon solid here as a reactant. So that means we need to reverse this partial reaction here. So when I reverse it, we're gonna say now we have 2 moles of H₂O gas, plus 1 mole of carbon solid, gives us 2 moles of hydrogen gas plus 1 mole of carbon dioxide gas. Remember, when you reverse the reaction that gives you the inverse of your k. So we're gonna get k is now to the negative one, so that's 2.75 times 10 to the negative one raised to the negative one. So our new K value here would be 3.6363. Now we do the same now we look at the next compound. So the next compound we're looking for is CO₂. If we look at the 2 partial reactions here, CO₂ here and here's CO₂ here. Remember, if we find it in more than one reaction we skip it. Next, we need 2 moles of carbon dioxide. So if we look, we're going to say we have 1 mole of carbon dioxide here. In the overall reaction there are 2 moles, but over here there's only 1 mole. That means I need to multiply this whole thing times 2. And actually, this equation is no longer relevant because it's been changed to the one in blue here. Alright, so I'm multiplying everything times 2 so I'm gonna have 2 H₂ gas, plus 2 CO₂ gas, gives me 2 H₂O gas, plus 2 CO gas. If we multiply the equation by 2, that becomes our new exponent or power for our K. So it's going to be k squared. So we're just going to square this value here. When I square it it gives me a new value of 0.585225. Step 3, you're gonna combine the partial reaction and cross out reaction intermediates if present. So remember, what's a reaction intermediate again? Well, these are compounds that look the same with one as a reactant and the other as a product, They must always cancel out. So if we go back up here, here are our 2 new reactions in blue. These former ones are gone now, they've been manipulated and changed. So what can we cancel out? What are intermediates? Well, we have 2 moles of H₂ as a reactant here, which cancel out with 2 moles of H₂O that are products here. What else? We have 2 moles of water as a reactant, 2 moles of water as a gas. Those cancel out. Anything else cancels out from here? Yes. We have one mole of CO₂ as a product canceling out with 1 from here that's a reactant, leaving us with one left. Now, if we look to see, nothing else can cancel out. So what's left at the end? We have 1 mole of CO₂ as a reactant, 1 mole of carbon solid as a reactant, 2 moles of carbon monoxide as a product. If you were to bring down everything, you would see that you just recreated your overall reaction. So we know we've done this correctly, and because we've recreated our overall reaction we just do step 4 which is where we multiply all the equilibrium constants of our partial reactions to obtain the overall equilibrium constant, of the overall reaction. So basically we're gonna say K_eq in orange, let's just do it in right here, K_eq in orange equals the new K_eq's that we've gotten. So remember, what are the new K_eq's? We have this new K_eq which we got from squaring the original K_eq, and then we got this K_eq from doing the inverse. We're just going to multiply those two values together to get our new overall K_eq. So plug that in, so we're gonna have 0.585225 times 3.6363, so that gives me 2.128 as my new K_eq of the overall reaction. And let's see, we're gonna round that to 3 sig figs so that gives me 2.13. So this would be our final answer.

Calculate K_c for: C(s) + ½ O₂(g) + H₂(g) ⇌ ½ CH₃OH(g) + ½ CO(g)  K_c = ?

Given the following reactions:
4 C(s) + 2 O₂(g) ⇌ 4 CO(g)                               K_p = 2.11 x 10⁴³
CH₃OH(g) + H₂O(g) ⇌ CO₂(g) + 3 H₂(g)          K_p = 7.17 x 10^–2
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)                     K_p = 2.00 x 10³

• A. 1.13×10¹³
• B. 5.60×10¹³
• C. 1.37×10¹⁵
• D. 5.35×10¹³

Here’s what students ask on this topic:

When we add two reactions together, how can we determine the value of the equilibrium constant for the new reaction?

When you add two chemical reactions together to form a new reaction, the equilibrium constant for the new reaction can be determined by multiplying the equilibrium constants of the original reactions. This is because the equilibrium constant, denoted as K, is a reflection of the ratio of the concentrations of the products to reactants, each raised to the power of their stoichiometric coefficients.

For example, if you have two reactions:

1. A → B with equilibrium constant K₁
2. B → C with equilibrium constant K₂

And you add them to get the new reaction:

A → C

The equilibrium constant for this new reaction, K₃, is the product of K₁ and K₂:

K₃ = K₁ * K₂

This relationship holds because the equilibrium constant is derived from the rate constants of the forward and reverse reactions, and when reactions are added, their rate constants multiply in a way that leads to the multiplication of their equilibrium constants. Remember, this only applies if the reactions are added in such a way that intermediates like B in this example are the same (i.e., they cancel out).

The table above provides the Keq values for two reactions. What is the correct mathematical expression needed to determine the equilibrium constant of the reaction shown below?

To answer your question, I need to clarify that the table mentioned is not visible in your query. However, I can guide you on how to determine the equilibrium constant (K_eq) for a given reaction using the K_eq values of other reactions.

If you have two reactions:

1. A -> B with K_eq = K_eq1
2. C -> D with K_eq = K_eq2

And you want to determine the K_eq for a new reaction that is a combination of the two, such as:

A + C -> B + D

You would use the following mathematical expression:

$K{\mathrm{eq}}_{\text{new reaction}}=K{\mathrm{eq}}_1\cdot K{\mathrm{eq}}_2$

This is because the equilibrium constants of individual reactions multiply when reactions are added together.

If the new reaction is the reverse of one of the given reactions, you would take the reciprocal of the K_eq value for that reaction. If the stoichiometry of a reaction is multiplied by a coefficient, you would raise the K_eq to the power of that coefficient.

For example, if the new reaction is the reverse of Reaction 1, the expression would be:

$K{\mathrm{eq}}_{}$