The following reaction of A3 molecules is spontaneous.

(b) What are the signs of ∆H, ∆S, and ∆G for the reaction? Explain.

Video transcript

Hey everyone, we're asked to consider the following reaction of B two molecules. If the reaction is spontaneous determine the sign of delta G, delta S and delta H. So looking at this reaction, we can see that we have B two molecules in our react inside and be three molecules in our product side. Now, to balance this out, we need to add a coefficient of three prior to our B two and a coefficient of two prior to our B three. Now let's go ahead and determine our signs. So we were told the reaction is spontaneous. So that means our delta G, which is the change in free energy must be negative since the reaction is spontaneous. Now let's go ahead and look at our change in entropy, which is our delta S. So looking at this reaction, we can see that our molecules decreased. So when the number of molecules decreases the collisions between them also decreases, resulting in a decrease in dis orderliness in the system. So this means our entropy is also going to be negative. Now let's go ahead and determine our NLP change which is our delta H. Now we know that our gibbs free energy equation is delta G equals delta H minus temperature times delta S. And in order for our delta G to be negative and our delta S to be negative. This means that our delta H must be negative as well. So our answer for our delta age is going to be negative based on our formula and also we can see that the process involves bond formation which solidifies our decision in saying that Delta H. Is negative. So these are going to be our final answers now. I hope that made sense and let us know if you have any questions.