Consider the following hypothetical reactions:
A ¡B HI = +60 kJ
B...

Question

Consider the following hypothetical reactions:
A ¡B HI = +60 kJ
B ¡ C HII = -90 kJ
(b) Construct an enthalpy diagram for substances A, B, and
C, and show how Hess's law applies.

Accepted Answer

welcome back everyone in this example we need to use his law to draw a diagram or an entropy diagram specifically for these substances present in the below reactions. So for our first given reaction we have why acting as a re agent. And this produces Z as a product. Were given a entropy value which corresponds to negative 120 kg joules. And then in our second reaction given from the prompt we have Z. Which now is being consumed as a reactant to form X. As a product corresponding to an entropy value of negative 70 kg joules. So we should recall first that delta H, which is our value for N. Therapy corresponds to the heat energy that is absorbed or evolved in a reaction. Now for us to tell whether the heat energy here is absorbed or evolved. We want to recognize that our entropy is negative for both reactions. And because entropy is negative, we can say that therefore, we can say that energy or he is released, meaning that we have an exhaust thermic reaction here. So we have two eggs. A thermic reactions now are prompt states that we should use Hess's law when drawing our diagram for our final answer. And we should recall that according to his law. If a reaction can take place By more than one route and initial and final conditions remain the same, then our total entropy change will remain the same. So we need to use a diagram to depict what is going on according to Hess's law with our given reaction. And as we can see here we have Z being produced as a product in the first given reaction where it's done being consumed as a reactant in the second reaction. And ultimately this means that because Z is both produced as a product in the first given reaction and then consumed as a reactant In the second given reaction, we can say that therefore Z is an intermediate. And so according to our second reaction, X is our final product. So in our diagram before we can actually draw the diagram, we want to write out a third reaction for the formation of why as a reactant to form X as the final product. And we should recall that using his law, this would be another route of getting to this product. And this means that our total change in entropy should remain the same. So we would say that our change in entropy is going to equal the sum from our first entropy of our reaction being negative 120 kg joules added to the sum of the entropy from the second given reaction in the prompt being negative kg joules. And so taking the sum here we get a change in entropy equal to a value of negative 190 kg joules. And so as we've outlined according to Hess's law above the total change in entropy going from reactant Y. Two product X is equal to a value of negative 190 kg joules. And this total change in entropy remains the same. Whether we go from reactant Y to form our intermediate Z. Corresponding to the change in entropy given in the prompt as negative 120 kg joules. Which is then added to the change in entropy going from Z as a reactant to our final product X. Which corresponded to the change in entropy equal to negative 70 kg joules. As given in the prompt. And so this diagram here would be our final answer Hess's law according to the given substances in our reactions. And so this would correspond to choice D in our multiple choice as our final answer. So I hope that everything that I reviewed was clear. But if you have any questions just lead them down below and I will see everyone in the next practice video.

Consider the following hypothetical reactions: A ¡B HI = +60 kJ B ¡ C HII = -90 kJ (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

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