Calculating K For Overall Reaction - Video Tutorials & Practice Problems

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 keq and orange 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 one mole of carbon solid reacting with one mole of carbon dioxide gas to produce two 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. All right. So how do we approach something like this? All we have to do is follow the steps provided before us. So step one 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 reactive, there's one 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. Uh 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 with the one from the overall reaction, right? So both of them are one mole, one mole that's good and they're both the same face solid and solid. It's important to remember the face isn't mean to match because sometimes you'll have water as a liquid in one partial reaction and water as a gas in another partial reaction, right? This may require you to reverse multiply or divide the partial pressures which will affect your equilibrium constant in blue. OK. Now step two, you keep moving on to the next compound in the overall reaction until you locate all compounds of the partial reactions, skip compounds found in multiple reactions. All right. So let's just do what it says there. We need to follow the overall reaction. So we need one 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 two moles of h2o gas plus one mole of carbon solid gives us the two moles of hydrogen gas plus one mole of carbon dioxide gas. Now 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, the two partial reactions, here's CO2 here and here's CO2 here. Remember if we find it in more than one reaction, we skip it. Next, we need two moles of carbon dioxide. So if we look, we're gonna say we have one mole of carbon dioxide here in the overall reaction, there's two moles, but over here, there's only one mole that means I need to multiply this whole thing times two. And actually, this equation is no longer relevant because it's been changed to the one in blue here. All right. So I'm multiplying everything times two. So I'm gonna have two H two gas plus 2 CO2 gas gives me 2 H2O gas plus two co gas. If we multiply the equation by two, that becomes our new exponent, our power for our K. So it's gonna be K squared. So we're just gonna square this value here. When I square it, it gives me a new value of 0.585225 step three, 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 two new reactions in blue, these former ones are gone now, they've been manipulated and changed. So what can we cancel out? What's what are intermediates? Well, we have two moles of H two as, as reacted here, which cancel out with two moles of H2O that are products here. What else? We have two moles of water as a reactant, two moles of water as a gas. Those cancel out anything else canceled out from here? Uh Yes, we have one mole of the CO2 as a product canceling out with one from here. That's reactant leaving us with one left. Now, if we look the seed, nothing else can cancel out. So what's left at the end? We have one mole of CO2 as reactant, one mole of carbon solid as a reactant, two 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 four, which is where we multiply all the equilibrium constants of our partial reactions to obtain the overall equilibrium constant uh of the overall reaction. So basically, we're gonna say KQ in orange, let's just do them right here. KQ and orange equals the new keqs that we've got. So remember what are the new Kaqs? We have this new Kaq which we got from squaring the original Kaq and then we got this KEQ from doing the inverse. We're just gonna multiply those two values together to get our new overall keq. So plug that in. So we're gonna have 0.585225 times 3.6363. So that gives me 2.128 as my new keq of the overall reaction. And let's see, we're gonna round that to 36 F. 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