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JF

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Problem 87

Among the many complex ions of cobalt are the following:

$$\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}(a q)+3 \mathrm{en}(a q) \rightleftharpoons \mathrm{Co}(\mathrm{en})_{3}^{3+}(a q)+6 \mathrm{NH}_{3}(a q)$$

where "en" stands for ethylenediamine, $\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2} .$ Six$\mathrm{Co}-\mathrm{N}$ bonds are broken and six $\mathrm{Co}-\mathrm{N}$ bonds are formed in this reaction, so $\Delta H_{\mathrm{rn}}^{\circ} \approx 0 ;$ yet $K>1 .$ What are the signs of $\Delta S^{\circ}$ and

$\Delta G^{\circ} ?$ What drives the reaction?

Answer

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## Discussion

## Video Transcript

okay, It's start this problem. We want to first right out. Um, the equations that we're gonna need So we're gonna need Delta G equals Delta H minus T Delta ass. And the question asks us to find the sign of Delta G and felt s. And we know that the K E Q is greater than one, so we can start with Delta G. So because we know the K Q is greater than one, this means that the products are going to be favored. This means of the reaction is proceeding to the right. Um, a better way to think about this is if we write out the que que que que expression right here. If K Q is greater than one, it means that C and D up here on the top has to be greater than a and B on the bottom to give a positive number. Oops, sorry. My mouse is being funky. So this means that the concentrate the concentrations of C Andy are greater than the concentrations of A and B. So the reaction proceeds to the right. So this is pretty much another way of saying that the reaction is spontaneous. So it doesn't need any input. Her energy, Um, because it's already perceiving to the right. It's already making the products. So because we know that the reaction is spontaneous, we know that the Delta G is going to be negative two. Okay, this is because for all reactions, you should definitely remember this all reactions a negative Delta G. It means the reaction is spontaneous, and it proceeds to the right. So when you see a K Q, that is greater than one, you can always assume that the Delta G will be negative. All right, so we know that we have to have negative Delta G. I'm just scroll down a little bit, rewrite our equation here, Felt G equals lips. Delta age minus t Don't s okay, so we know that this is negative and we're told that Delta H is zero. So we want to make sure that Delta s is signed will make Delta G negative. So let's just think first. What if Delta s was negative? Well, this wouldn't work, because we would be, um, subtracting a negative, which would cause, um, us to add to zero, which would make a positive value. Right? So That means that Delta s has to be positive because no matter how big or small this number gets, as long as it's positive, you will always be subtracting in causing a zero for Delta G. All right, And so the last part of the question Axa asks us What? What? What's causing the reaction? Teoh go forward. So we know it's spontaneous because we know the Delta G is negative. But what's driving the reaction? So we know the belt that s we just figured out is positive. And this means that entropy is increasing in the universe tends to want to increase entropy. So entropy increasing is driving this reaction.