Ethene (C₂H₄) reacts with halogens (X₂) by the following reaction:
C₂H₄(𝑔) + X₂(𝑔) ⇌ C₂H₄X₂(𝑔)
The following figures represent the concentrations at equilibrium at the same temperature when X₂ is Cl₂ (green), Br₂ (brown), and I₂ (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3]

Question

Ethene (C₂H₄) reacts with halogens (X₂) by the following reaction:

C₂H₄(𝑔) + X₂(𝑔) ⇌ C₂H₄X₂(𝑔)

The following figures represent the concentrations at equilibrium at the same temperature when X₂ is Cl₂ (green), Br₂ (brown), and I₂ (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3]

Accepted Answer

Welcome back everyone consider the following reaction when propane reacts with allergens. So we have an equilibrium reaction given here. We're told that the equilibrium concentrations at the same temperature when X two is iodine, purple chlorine, greene and Broome in red. Are illustrated in the following figures arranged the equilibrium from largest to smallest equilibrium constant. Let's begin by recalling that our equilibrium constant is represented by the term K. And it's equal to our concentration of products over our concentration of react ints. And based on our given reaction, we have our products which is our propane, ated or hydrogenated propane C three H six X two. And this concentration is going to be placed over our concentration of our reactant in which we have our halogen gas reactant and then our propane C three H six gas. So now with this equilibrium constant expression outlined, we can count the molecules in each of our illustrations to determine The value of K. Beginning with Figure one. We can count a total for our riding purple molecules As two molecules here. So our concentration of I two equals two molecules. Next let's count our concentration of our reactant, propane, so or propane C. Three H six. So counting the molecules of propane, we have 12 as well. So two molecules, it's the concentration here and then for our concentration of our product which is C three H six X two which is our originated propane, we have a total in illustration one being 1234567 molecules. And so writing out our equilibrium constant expression and plugging everything in accordingly we have our concentration of our product which we can plug in as seven divided by our concentration of our reactant. Sorry this should be a division symbol. So our first reactant we have is iodine, which we counted two molecules as the concentration for, multiplied by our concentration of our second reactant, propane which we have a concentration of two as well. In illustration one simplifying this expression, we will find an equilibrium constant equal to a value of 1.75. And so now let's move on to illustration to to sulfur K. So noting down and according to illustration to we're focusing on our halogen now being chlorine as the green molecules. So noting down our concentration of chlorine In illustration to counting the green molecules, we have 123. So that is our concentration is three molecules. Then we have our concentration again of propane C three H six. In illustration to will see a total concentration being molecules of propane and then our concentration of our products C three H six X two. And rather instead of X two, we can fill in cl two. So we should actually do that for Illustration one as well. We can make the formula C three H six I two. Anyways counting our concentration of C three H six cl two in illustration to we have molecules here. And that is our concentration. And so counting our or calculating our equilibrium constant K. We have our concentration of our product seven molecules divided by our concentration of our reactant cl 23 molecules and propane. Three molecules that's seven divided by three times three. And this quotient gives us a result equal to the value of 0.77 Which we can round to about 0.78 since it's repeating sevens so so far based on our calculated the equilibrium constants, we can rank illustration one as being the largest so far. But let's consider illustration three. Next beginning with counting our concentration of our third halogen which is mentioned to be bro mean. So this is b. r. two for our concentration of grooming. In illustration three we can count a total of just one bromine molecule since it's the red molecules, then our concentration of propane C. Three H six. In illustration three counting our molecules, we have one as well And then our concentration of our products which would be c. three Age six b. r. two. We would count a total of 12345678 molecules total. And writing out our equilibrium constant expression, we have K equal to our concentration of our product C. Three H six pr to which we set as eight divided by our concentration of our and we can use parentheses actually divided by a concentration of our first reactant roaming which we counted a concentration of one molecule and then our second reacting propane. We also counted one molecule that's eight divided by one, Giving us a equilibrium constant value of eight. And so writing out our ranking for each of our equilibrium, we would say that from largest to smallest, the largest equilibrium is associated with Illustration three, in which we had a K value of eight which was greater than our Equilibrium. for illustration one being 1.75 And illustration one had a greater equilibrium than illustration to which only had a equilibrium constant value of .78. And so for our final answer, we've correctly outlined our ranking of the equilibrium for each of our illustrations and so our final answer is going to correspond to choice a in the multiple choice as the correct answer. I hope this made sense. And let us know if you have any questions.

Verified Solution

Key Concepts

Equilibrium Constant (K)

Le Chatelier's Principle

Halogen Reactivity