Using the bond-dissociation energies in Table 5.6 (see Section...

Question

Using the bond-dissociation energies in Table 5.6 (see Section 5.3.1), estimate the equilibrium constant of the following reaction at 298 K.

Accepted Answer

All right. Hi everyone. So this question is asking us to use the bond dissociation energies to estimate the equilibrium constant of the reaction given below at 298 Kelvin. So here we have a chemical reaction with relevant bond association energies and units of kilojoules per mole. And four different answer choices for the value of K eq or the equilibrium constant. Now to understand the way in which we would solve this equation recall that the equilibrium constant can be calculated using our delta G standard or jibs free energy. Specifically delta G standard is equal to negative RT multiplied by the natural logarithm of the equilibrium constant. But the problem is that we do not quite know what Dulce G standard happens to be. So we can use a separate equation to figure out what delta G standard is before we can solve for keq right. Delta G standard is also equal to delta H standard subtracted by T delta S standard. So using the bond association energies, we can calculate delta H standard or the entropy of this reaction and therefore find the jibs free energy. Now notice also that the entropy or delta S standard is also a term in this equation recall that entropy refers to the degree of disorder or randomness within a given system. And depending on whether or not there's a change in the number of molecules from the reactant side to the product side, an entropy can either increase, decrease or stay the same. Now, on the reactant side, we have two different molecules. One of them is an all keen and the other is BR two. And then on the product side, we still have two molecules, one is an aloe bromide and the other is H pr So the main idea here is that because there's no change in the number of molecules from the reactant side to the product side, entropy has not actually changed. So therefore, delta S standard, in this case is equal to zero. And this essentially eliminates the second term of our equation here for delta G standard. What this means is that delta G standard is equal to delta H standard. So let's go ahead and figure out what delta H standard it's and therefore, what delta G standard is, we can calculate the value for the delta H standard by taking the sum of all of the bond association energies that are relevant for your reaction. Now, it's worth mentioning that all of the bond association energies provided are given as positive values. But that may not always be the case. Recall that when a particular bond is formed, right? Then the value of the entropy is going to be positive or greater than zero correction. Sorry about that. But what I meant to say is that if a particular bond is formed, then the entropy is going to be negative. Whereas if it's broken, then the entropy is going to be positive. So what I'm going to do here is I'm going to discuss very briefly what bonds have been broken and which have been formed and then place the appropriate sign in my table to make sure I do not lose track. Now, as for the bonds broken, we can see here that one of the hydrogens on the aic carbon of my alkene has been replaced by a bromine. So this means that a carbon hydrogen bond has been broken and a carbon bromine bond has been formed. So the bond association energy of the carbon bromine bond should be negative because it's been formed. Whereas the value of the carbon hydrogen bond dissociation energy should remain positive. And also the bond between the two brom means of BR two must have been broken to facilitate its conversion into HBR right. So the bromine bromine bond has been broken, meaning it should stay positive and the hydrogen bromine bond has been formed. So it should be negative. So now that we have the appropriate sign, we can go ahead and calculate the sum of all of our bond association energies and therefore calculate the entropy So we here, that would be 410 kilo jules per month, add it to 192 kg joules per mole, subtracted by 285 kg joules per mole and also subtracted by 368 kilo jules Perma. And so when I combine all of my values together, I get that delta H standard and therefore delta G standard are equal to negative 51 kilojoules perm. So now we can go ahead and estimate the value of the equilibrium constant. So let me scroll down to do that. Once again, recall that delta G standard is equal to negative RT multiplied by the natural logarithm of the equilibrium constant. Now R in this case refers to the gas constant which is 8.314 joules per mole, Kelvin. But theres a slight problem, right? Because the value of delta G standard has been calculated in kilojoules per mole. So to fix this discrepancy, I'm going to go ahead and take the value of delta G standard in kilojoules per mole and multiply this by 1000 because recall that there are 1000 jewels and 1 kg jewel. So when I do this, I get that delta G standard is equal to negative 51,000 jules promo. And now all of my units are consistent. And t for the record refers to temperature in units of Kelvin. An earlier temperature was provided to us which was 298 Kelvin. So now we can make our substitutions in my equation. Delta G standard is negative 51,000 jules perm. And that's equal to negative R. That's 8.314 Jules promo Kelvin multiplied by temperature which is 298 Kelvin multiplied by the natural logarithm of the equilibrium constant. So the goal here is to isolate the natural logarithm of K eq and make sure it is by itself on one side of the equation. So to do this, I would have to subtract or sorry, divide both sides by negative RT. Let me actually scroll down just a little bit more. So here I would have to divide both sides by negative 8.314 Jules Promo Kelvin multiplied by 298 Kelvin. And as it turns out, this actually cancels all of my units present in the equation. So that the natural logarithm of keq is equal to negative 51,000 divided by negative 8.314 multiplied by 298. And this gives me a value of 20 point 584 67. So now to solve for the equilibrium constant, I get that K EQ is equal to E to the power of 20.58467 which gives us our final answer of 8.7 multiplied by 10 to the power of nine. And there you have it. Here is our final answer and this corresponds to option A in the multiple choice. So KQ for this reaction is equal to 8.7 multiplied by 10 to the power of nine. So there you have it. If you stuck around until the end of this video. Thank you so very much for watching and I hope you found this helpful.

Using the bond-dissociation energies in Table 5.6 (see Section 5.3.1), estimate the equilibrium constant of the following reaction at 298 K.

Key Concepts

Bond Dissociation Energy

Equilibrium Constant (K)

Thermodynamic Relationships

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