Skip to main content
Ch.19 - Chemical Thermodynamics

Chapter 19, Problem 9a

Consider a reaction A2(𝑔) + B2(𝑔) β‡Œ 2 AB(𝑔), atoms of A shown in red in the diagram and atoms of B shown in blue. (a) If 𝐾𝑐 = 1, which box represents the system at equilibrium?

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
317
views
Was this helpful?

Video transcript

Hello everyone. So in this video we're given three different diagrams for this reaction right over here and we're trying to see which box represents the system at equilibrium. So if our constant is equal to one. So we're given these three boxes and these three diagrams represent different mixtures of reacting and products for our reaction. So let's go ahead around the equation to get the equilibrium constant, that's going to be equal to the concentration or the molds of our products over our reactivates in our case, then we have our products being a B to the power of three because we have a coefficient of three in our reaction, The concentration or moles of our a. three multiplied by the concentration or moles of B. three. Okay, so determine which system are which box represents system at equilibrium if our reaction quotient is equal to one, which is denoted as Q. So Q. Is equal to one, the system is at equilibrium. So let's go ahead and start off with our box number one. Let's go ahead and count how many of our products that we have. So that's 123. So we have three to the third power. And let's see for R. A. We have 12 and three. And then for B we have 12 and three. So three multiplied by three. So you put this into the calculator to get the value of three. Now doing this, but for box two in the middle let's see here for our products, we have one and two and then for our A we have 123 and four. And then for B. R. Reds is just one and two. Again, putting this into the calculator, we get a value of one. Now, lastly for box three, our last box here again for our products we have one and two. Now for our A. And B. For A. We have 12 and three. And for our B. We have 123 and four. Alright, putting this into the calculator, I'm just gonna go ahead and round this will be 0.67. So again, we want the reaction reaction Kocian to be equal to one. In this case only box to has that if our Q is equal to one then the system is at equilibrium. So therefore, let's put this in writing our answer is that Box two is in equilibrium and this right here is going to be my final answer for this problem. Thank you all so much for watching