Consider a reaction A₂(𝑔) + B₂(𝑔) ⇌ 2 AB(𝑔), atoms of A shown in red in the diagram and atoms of B shown in blue. (b) If 𝐾_𝑐 = 1, which box represents the system at 𝑄 < 𝐾_𝑐?

Video transcript

well everyone. So in this video we're trying to see out of all the three boxes given here which one represents a system where the queue is less than our constant Q. Represents our reaction quotient. So these diagrams right here represent three different mixtures of our reactions and products for are given chemical reaction right over here. So let's first write out the equation to get our equilibrium constant. That is of course equal to K. C. Equaling to the products or the concentration or motor products over the molar concentration of our reactant. So using our chemical reaction for my products we have a. B to the power of three. Of course because we have the coefficient of three in our chemical reaction. Then for our reactant since our starting region. So we have a. three and B. Three. Alright, so let's go ahead and calculate for Q. The reaction quotient for H. Box. So starting off with box one. For our products We have 1, 2, 3 moles. So that be 3 to the three. And then for our a. We have three moles as well for our B. So putting that value into my calculator, I get the Q. equaling 2 3. The same exact process for a box too. For our products we have 1, 2. Then for my A. I have 123 and four. And then for my B. We have one and two And putting this into my calculator, I get the q value of one And lastly our box number three here For our products, we have one and for my A. We have 12 and three and four R. B. We have 123 and four. Again putting this into my calculator. I'm just gonna go ahead and round to two decimal places, giving me a value of 0.67. So again we want our quotient to be less than our KC value. KC value is equal to one, and box three represents that Requirement. And therefore putting our answer into writing is that our box three has a reaction quotient less than our constant. Right over here are K C value. So therefore this is going to be my final answer for this problem.