The dehydrogenation of butane to trans-but-2-ene has ΔH° = +116 k...

Question

The dehydrogenation of butane to trans-but-2-ene has ΔH° = +116 kJ/mol (+27.6 kcal/mol) and ΔS° = +117J/kelvin-mol (+28.0 cal/kelvin-mol). a. Compute the value of ΔG° for dehydrogenation at room temperature (25 °C or 298 °K). Is dehydrogenation favored or disfavored?HINT: When you are doing synthesis problems, avoid using these high-temperature industrial methods. They require specialized equipment, and they produce variable mixtures of products.

Accepted Answer

Welcome back everyone. The d hydrogenation of methane to ethane has an entropy change of positive 1 36 kg per mole. Also equivalent to positive 32.5 kilocalories per mole and an entropy change equal to positive 1 20 joules per kelvin mole, also equivalent to 28.7 calories per kelvin mole, calculate the value of gibbs free energy for the reaction at room temperature. Is the reaction favored or disfavored? We're going to begin by recalling that when we have a gift free energy change for the reaction that is greater than zero. That means the reaction is non spontaneous and not favored. But when we have a gibbs free energy change that is less than zero. That means the reaction is spontaneous and therefore favored in the forward direction. And so with this outlined, we also want to write out our reaction. So we have ethane, Two carbons bonded to each other where each carbon is surrounded by three hydrogen atoms. And we want to recall that an A. D hydrogenation and sorry, this is a hydrogen here. But in a D hydrogenation reaction, we're going to take our reactant and react it with a solid catalyst. So we can use something like palladium bonded to carbon as our catalyst. Where in this type of reaction, we have the removal of a hydrogen from each carbon in the molecule and so we would remove two hydrogen is here where our product is going to result in a product with a double bond and hydrogen gas as our second product. So this would form a thin as our first product where we still have two carbons but they're now bonded to one another in a double bond. So we have a pi bond. And now each of these carbons only have two bonds to hydrogen. And we should recognize that carbon should have a double bond between one another. Because we still want carbon to be stable with four bonds. And making a pi bond between these two carbons allows each carbon to have a total of four bonds Again, as we stated, we have hydrogen gas as our second product And this completes our d hydrogenation reaction here. Now that we have our reaction outlined, we want to calculate our gibbs free energy change where we should recall. We can calculate it by taking our Topic change given in the prompt and subtracting it from the temperature of the reaction times the change in entropy of the reaction. So now we just want to plug in our knowns based on what we have given from the prompt for our entropy value were given a value of positive 136 killed Jules Permal which we're subtracting from the temperature which were not given in the prompt but we're going to assume standard temperature being at 25 degrees Celsius. Which we need to convert to kelvin by adding to 73.15. And this is going to give us a kelvin temperature of 2 98.15 kelvin. And so we're subtracting from 2 98.15 kelvin and multiplying by our Entropy change given in the prompt as 120 jewels per kelvin mole. Which we need to convert to kill a jewels so that we can subtract from our initial value being our entropy in the beginning of the equation here. So we're going to multiply by a conversion factor where we have jewels in the denominator and kilo jewels in the numerator where are prefix kilo and this is a one here are prefix kilo, tells us that we have 10 to the third power of our base unit jewels. And so this allows us to cancel out jules leaving us with kilo joules per kelvin mole. And now in our next line we can simplify so that we have 136 kg jewels Permal subtracted from 2 98 kelvin to 98.15 kelvin multiplied by 0.1 to zero killed joules per kelvin mole. And so now simplifying this further, we're going to have 136 kg jewels Permal subtracted from the result of our product which is going to be 35.778 we can cancel out our units of kelvin and we're left with units of kilo jewels per mole. And so finally taking this difference here, we have a result of 100.2 kg joules per mole which we can round 23 sig figs as or one sig fig as just 100 kg joules per mole. And let's fix that killer jewels Permal. And we want to recall that our conversion factor to go from kila jewels to kill a cow's where we have for one kill a cow 4.184 kg jewels. And so this will also result in a value of about 23.9 kg cows Permal. Because we'll be able to cancel out our units of killing jewels. Now we want to recognize that both these values are positive values. We have positive 100 kg per mole and positive 23. kg per mole. And so therefore our our gifts. Free energy value is greater than zero because their positive values. And so the d hydra de hydrogenation is not favored at room temperature. And so this will be our final answer to complete this example, which corresponds to choice B as the correct answer in the multiple choice. I hope everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

Key Concepts

Gibbs Free Energy (ΔG°)

Enthalpy (ΔH°)

Entropy (ΔS°)

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