For the reaction shown here, Kc = 0.513 at 500 K. N2O4(g) ⇌ 2NO2(...

Question

For the reaction shown here, Kc = 0.513 at 500 K. N2O4(g) ⇌ 2NO2(g)
If a reaction vessel initially contains an N2O4 concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N2O4 and NO2 at 500 K?

Accepted Answer

hi everyone for this problem, we're told to consider the following hypothetical equilibrium reaction. We need to determine the concentration of A. And B. Once equilibrium is established, assuming you start only with 2.50 molar of A. So we're going to calculate the equilibrium concentrations from the given initial concentration and equilibrium constant using an ice table. So let's go ahead and rewrite out our equation. So we have to a and we have to be okay, so we ride out ice on the side and I'm going to draw this line here to separate our products from our reactant, since we are told we have 2.5 molar of A. So this is what we start off with initially, this is what our I wrote of our ice table represents. And so at the very beginning of the reaction, we have 2.50 moles of A. And we have zero moles of of B. So the reaction is currently not at equilibrium because we have that zero mol of product. So our reaction is going to shift in the forward direction in order to reach equilibrium. So what that means is our reactant is going to be consumed and our products is going to be created. So, in terms of R C ro for our ice table, our change in concentration for our reactant is going to be negative and our change in concentration of reactant is going to be positive. So now we need to pay attention to our stoke eom a tree. So we have two moles of A. So we're going to have minus two X. And we also have two moles of B. So this is going to be plus two X. Ok. And r E is our equilibrium concentration, which is what we're trying to solve for. So we're going to add up our first two rows and just write it out here. So we have 2.50 minus two X. And then we just have to X here. Alright, so we need to solve for X in order to figure out what our equilibrium concentration is going to be for both A and B. For us to solve for X. We're going to need to write out our equilibrium expression and that expression is K C is equal to our concentration of products over our concentration of reactant by doing this, we're going to be able to solve for X. And the problem, they tell us what our K C value is, it's 5.64. So let's go ahead and plug that in here. Our concentration of products, we see here that it's two X. Okay, so we have two X. But we need to raise it to its Tokyo metric coefficient because we have two moles of B. We need to raise this to the second power for our concentration of reactant, we have 2.50 minus two X. Okay, we're copying everything from our equilibrium row here for this concentration of products and concentration of reactant. So we have 2.50 minus two X. And we need to pay attention to the stoke eo metric coefficient because there's a two there, we need to raise this to the second power. Okay, so our goal here is to solve for X and by solving for X will be able to plug it into our equilibrium row and figure out what our concentrations are. So what we can do here because our rights, everything is square a ble here, we can take the square root of both sides of our equation. And when we do that we get rid of the squares on the right side. So we're going to take the square root of both sides. And when we do that we get a simplification of two point 3749 is equal to two X over 2. minus two X. Okay, so we're going to multiply both sides by 2.50 minus two X. So we have 2. times 2.50 minus two X is equal to two X. So our goal is to solve for X. So let's go ahead and foil out. When we foil out we get 5. minus 4. X is equal to two X. We can divide both sides by let's simplify some more. So we get 5.9471 is equal to 6. 97 X. So we can divide both sides by 6.7497. And when we do that we get X. Is equal to 0.881. Okay, so now that we know what X is, we can plug in X to our equilibrium row of our ice table here to solve for the equilibrium concentration of A. And B. So let's go ahead and do a first. So are equilibrium concentration of A says that it is 2. minus two X. So we know what X is. We just solved for it. Let's bring it up here. We said X is equal to zero point 881. Okay, so that means that our concentration of A is going to equal 2. minus two times 0.881. Which gives us a concentration of 0. Moeller for A. And our equilibrium Rose says that our equilibrium concentration for B is going to equal to X. So we have two times 0. and this gives us a equilibrium concentration of 1.76 moller for B. So these are going to be our final answer. The concentration of A is 0.740 moller and our equilibrium concentration of B is 1.76 smaller. Okay, so that is the end of this problem. I hope this was helpful

For the reaction shown here, Kc = 0.513 at 500 K. N2O4(g) ⇌ 2NO2(g) If a reaction vessel initially contains an N2O4 concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N2O4 and NO2 at 500 K?

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