Calculate ∆G°, ∆H°, and ∆S° for the...

Question

Calculate ∆G°, ∆H°, and ∆S° for the following acid–base reactions. Rationalize the value of ∆H° based on the structure of the conjugate bases. [Assume T = 298 K.]

(c) Chemical equation showing the reaction between hydronium ion and bromide ion to form water and hydrobromic acid.

Accepted Answer

Welcome back everyone. What is the Standard Gibbs free energy change? Standard entropy change. So let's label those and standard entropy change for the provided acid based reaction explain the rationale behind the value of the entropy change by considering the structures of the conjugate base assume a temperature of 298. Kelvin. We're given the equilibrium reaction and we can immediately estimate the entropy change of the reaction which is about zero. And how do we know that? Well, essentially all of those states are aqueous, right? If all of those states are aqueous and we have the same number of the total moles on each side, specifically one plus one gives us two moles on the left, one plus one gives us two modes. On the right. We can say that there is nearly no change in the disorder of the system. So delta S is approximately zero based on the equation, delta G is equal to delta H minus C delta S. And since delta S is approximately zero, we can state that the Gibbs free energy change is approximately equal to the entropy change of the reaction. OK. So we have one of our conclusions, the entropy changes about zero. Now we're going to determine the gibbs free energy change, which is negative RT Ln of the equilibrium constant KEQ. It is important to understand that the equilibrium constant is going to be calculated from the equilibrium itself using the PK A values. So we can say negative RT Ln of what is the equilibrium constant keq? Well, essentially it stands to the power of negative PKEQ. The next question is what is PKQ? We have to recall that PKQ is the difference between the PK of the reactant acid and the PK A of the product asset. If we distribute the negative sign, we end up with negative RCLN. Sun raised the power of PK A of the product acid minus PK A of the reactant acid. And we can substitute the values of interest. First of all, let's plug in the value of R. We are going to add the negative sign. Then we're going to add one point 987 calories. I all poor Kelvin was the temperature 298 Kelvin. Now we're going to multiply by the natural logarithm of 10 rays to the power of PK A of the product acid. Well, essentially the product acid is HCL, right? Because in a reverse reaction, it can donate a proton to become chloride and the reactant acid is H to us because it donates a protein to become HS minus. So the product asset which is HCL has a PK value of negative seven. So we're going to put negative seven and subtract seven which is the PK A of H two S OK. And now let's perform the calculation specifically, we're going to get 19 0.1 kilo calories for more which is both the delta G and the delta H valley. So we can essentially remind that this is delta G and this is also approximately equal to delta H. And now that we have all of the required values, we want to try to understand the value of the entropy change, right? Explain the rationale behind the value of delta H by considering the structures of the conjugate base. Now, the conjugate bases are chloride and HS minus. Now considering the size because those are in the same period, sulfur and chlorine, right. Chlorine specifically is smaller because when we go from left to right across the period, the atomic size decreases and sulfur is larger. Besides what do we know about the electron negativities, it's more important to think about the electron negativities actually, right? Because they have relatively similar sizes, right? Meaning the most important factor is the electron negativity in the electron negativity of chlorine is greater than the electron negativity of sulfur. That's what we know because when we go from left to right, electron negativity increases. And this tells us that if chloride more specifically, if chlorine has a higher electron values, right, then it's keeping its electrons closer to itself in a formation of a covalent bond. And also we can state that chloride he is a more stable or less reactive then hs minus because it's more electron right. And if it's more electron, it means that it's not allowing its electrons to capture a proton that easily because those electrons are held tightly by the nucleus of chlorine. And this explains why the value of the entropy change is positive. Why we expect to have a shift to the left side? Because we always have a shift towards a we carry electrolyte. In this case, we have our explanation and the required values. So first of all, the entropy change is about zero, the entropy change is about 19.1 k calorie sperm, which is also the change in that gives free energy. And now the value of the entropy change is due to the conjugate base in the backward reaction being more stable since chlorine is more electron, right. And this essentially tells that the equilibrium shifts towards the formation of the reactants. Since delta G is also positive, that would be it. Thank you for watching.

Calculate ∆G°, ∆H°, and ∆S° for the following acid–base reactions. Rationalize the value of ∆H° based on the structure of the conjugate bases. [Assume T = 298 K.]
(c)

Key Concepts

Gibbs Free Energy (∆G°)

Enthalpy (∆H°)

Entropy (∆S°)

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