Formation constants for the ammonia and ethylenediamine complexes of nickel(II) indicate that Ni(en)₃²⁺ is much more

stable than Ni(NH₃)₆²⁺:

(1) <REACTION>

(2) <REACTION>

The enthalpy changes for the two reactions, ΔH°₁ and ΔH°₂, should be about the same because both complexes have six Ni﹣N bonds. 

(c) Assuming that ΔH°₂ - ΔH°₁ is zero, calculate the value of ΔS°₂ - ΔS°₁.

Video transcript

Hi, everybody. Welcome back. Let's take a look at our next problem. Consider the following formation constant for the complexes of a certain metal M. We have two equations labeled one and two. In equation one, we are starting with our metal M with six water ligands and a positive two charge in aqueous form, reacting with three molecules of en or ethylene diamine in aqueous form in a reversible reaction forming a complex of our metal with three ethylene diamine ligands and a positive two charge plus six molecules of water. And our KF one, our formation constant for this equation one is 8.0 times 10 to the 17th power. In equation number two, we start the same way with our M with six water ligands and a positive two charge plus six molecules of NH three ammonia. In reversible reaction forming a new complex of our metal with six ammonia ligands and a positive two charge plus six molecules of water. RKF two is 4.0 times 10 to the eighth power. The values indicate that M with ammonia ligands is less stable than M with ethylene diamine ligands because both reactions have six MN bonds and nitrogen bonds. The change in enthalpy for the two reactions. Delta H one and delta H two should be the same if delta H two minus delta H one is zero. What is the value of delta S one minus delta, delta S two, excuse me minus delta S one. Let's keep in mind that all of the delta H delta S are have that little circle symbol, meaning under standard temperature and pressure, our answer choices are a 3.0 times 10 to the 10th Joel per mole. Kelvin B 1.8 times 10 to the negative second joules per mole. Kelvin C negative 3.0 times 10 to the negative 10th. Joules per mole Kelvin or D negative 1.8 times 10 to the second joules per mole Kelvin. So as our question tells us, there will be no change in entropy because the entropy, when you're talking about the formation of something depends on the energy to break certain bonds and form other bonds. And in this case, the bonds that are breaking and formed are the same. You've broken six bonds to the water ligands, you've formed six bonds to nitrogen. So there is no change in enthalpy. So we want to focus on the change in entropy and the difference between the two reactions because that will determine which is more stable or less stable. So we have formation constants. We need to get to delta S. So we do have equations that can relate these two and that both can be related to delta G. So delta G standard is equal to negative RT Ln where Ln is natural log of K and R is the gas constant. So there comes our formation constant. And then we also have the equation of delta G standard is equal to delta H standard minus T delta S standard. So since both of these equal delta G, the change in gibbs free energy, we can set them equal to each other. However, we're not just looking for a delta S, we're looking for that value of delta S two minus delta S one. So we can get to this by focusing on that second equation and saying delta G. Now here, I'm going to say delta G one minus delta V two, even though my eventual goal is to have delta S two minus delta S one. Because when I look at that equation, there's a negative sign in front of the T delta S. So that's going to reverse the order of subtraction. So that's why I'm setting it that way. So delta G one minus delta G two, well, therefore equal delta H one minus T delta S one that in parentheses minus and then in parentheses, delta H two minus T delta S two. So you can see how in the end that order of subtraction is going to flip around due to our negative sign. And we can rearrange that equation by combining our delta HS delta H one minus delta H two. And then T delta S two is now positive. So we have plus T delta S two minus T delta S one. Well, we know that our delta H one minus delta H two must equal zero. It will cancel out. So we're left with T delta S two minus T delta S one. And of course, the T can pull out and we are left with T times delta S two or sorry, T multiplied by delta S two minus delta S one. So now we've got one of our terms here or we, as one of our terms, we have the value we're looking for. So let's set that equal to our expression with K in it. So we have T delta S two minus delta S one will be equal to. And now here again, we need to remember that we're looking at not just delta G but delta G one minus delta G two. And so it will equal negative RT Ln natural logarithm of K one minus negative RT LNK two. So we can see here where our temperatures are going to cancel out and we can combine these terms. And if we remember that when we're talking about natural logarithms, now put up at the top the rule, we remember that the rule that the natural logarithm of X minus the natural logarithm of Y is equal to the natural logarithm of X divided by Y since we're talking about exponents, when we talk about logarithms. So with that in mind, let's simplify our equation and we have that RT times delta S two minus delta S one equals. And now because we have that negative sign, we switch around our order of subtraction and we have RT LNK and I edited, I had forgotten the F after KF this is a formation constant. So we should have KF two minus RT LNKF one. We see how our temperatures now cancel out on both sides of the equation. And if we use our rule about natural logarithms, we can simplify that equation as delta S two minus delta S one. So the value we're looking for equals R times the natural logarithm of KF two divided by KF one. Well, now we have all the values we need to solve this equation. So we have our value of R the gas constant, which is 8.314 jules per mole coven and our natural logarithm. So multiplied by the natural logarithm of 4.0 times 10 to the eighth RKF two divided by 8.0 times 10 to the 17th RKF one. So we head to our calculator and do that multiplication. And we end up with uh rounded off negative 1.108. Excuse me, we end up with rounded off negative 182.4 joules per mole Kelvin, our answers are given in scientific notation, we have two significant figures. So that's going to end up giving us negative 1.8 times 10 to the second jewel per mole Kelvin. And we see that that answer is available here as choice. The started running out of space. So I scrolled up a little bit. So once again, we were given these formation constants and we used our equations to use gibbs free energy to relate our expression that includes delta S with our expression that includes our formation constant. And we've calculated that the value of delta S two minus delta S one is equal to choice D negative 1.8 times 10 to the second joule per mole. Kelvin. See you in the next video.