Consider the reaction and the associated equilibrium constant: aA(g) ⇌ bB(g) Kc = 4.0 Find the equilibrium concentrations of A and B for each value of a and b. Assume that the initial concentration of A in each case is 1.0 M and that no B is present at the beginning of the reaction. c. a=1;b=2

Video transcript

Welcome back everyone to another video, consider the reaction and the associated equilibrium constant, a moles of a gas in equilibrium with B moles of B gas. The equilibrium constant is 2.0. Find the equilibrium concentrations of A and B for A equals one and B equals two. Assume that the initial concentration of A in each case is 0.10 molar. And that no B is present at the beginning of the reaction. And we're given for answer choices. ABC and D they provide us with different concentrations of A AND B at equilibrium. We're going to solve this problem and identify the correct answer. So first of all, we're given our equilibrium constant KC. And what we can do is essentially express it in terms of the given concentrations, we take our products. So in this case, we have one product B, we need to raise it to the coefficient of lowercase B. And we need to divide that by the concentration of A raised to the power of lower case A, those concentrations are measured at equilibrium because we're looking at the equilibrium constant. Now, what we can do here is just rewrite the equation and we are going to set up an IC table. Now we have I initial C change and E equilibrium. What do we know? Well, essentially the problem states that the concentration of A is 0.7 molar. And initially, we have no B so the concentration of B initially is zero, let's assume that the change of A is negative A, right? Because that's the number of moles of A. But in addition to that, we need to use X specifically, it's not necessarily the coefficient itself, it's some amount of X. And if we use the geometry, we need to multiply the coefficient A by X. And similarly, for B, we're going to say the amount of B produced to B plus BX. So we're producing the same amount X multiplied by B due to stry. And our equilibrium the amounts would be 0.7 minus A X molar. So we can actually add our units everywhere. And for B, we would have BX molar. Therefore, we can say that the equilibrium constant KC is simply BXBX, raise the power of B divided by 0.7 minus A X, raise the power of A. Now we can use A and B values if B is two, then we have two X squared. So that'll be four X squared divided by 0.10 minus A is one. So minus X, the power of one, meaning we don't need to use pars, but actually let's use them based on the division sign that would be more accurate to say that we have 0.7 minus X in Brees. And we know that this value is equal to 2.0. So what we're going to do is just rearrange this equation. If we multiply both sides by the denominator, then we can say that four X squared is equal to 0.7 minus X multiplied by two, which gives us 0.20 minus 2.0 X. And now if we rearrange this into a quadratic equation, we get four X squared plus 2.0 X minus 0.20 equals zero. We need to apply the quadratic formula and find two solutions. We're going to take the one that is positive and the one that gives us positive equilibrium concentrations. So if we solve the equation, we end up with the value of X which is equal to zero point zero 8541, now what we're going to do here is just find the equilibrium concentrations. So we are going to start with a, a at equilibrium will be 0.7 minus A X. So essentially if A is a one, we're just subtracting X. So we're subtracting 0.08 541. This gives us the concentration of a 0.015 molar. And now what about the concentration of B at equilibrium? We have BX if B is two then we have two xo. So we are essentially taking two multiplied by 0.08541 molar, which essentially gives us 0.17 molar. Looking at the answer choices, we can sell that option. D is the correct answer to us, right? We have a at equilibrium being equal to 0.015 molar and B at equilibrium of 0.17 molar. That would be it for today. And thank you for watching.