For each reaction, calculate ΔH°_rxn, ΔS°_rxn, and ΔG°_rxnat 25 °C and state whether or not the reaction is spontaneous. If the reaction is not spontaneous, would a change in temperature make it spontaneous? If so, should the temperature be raised or lowered from 25 °C? a. 2 CH₄(g) → C₂H₆(g) + H₂(g)

Question

Accepted Answer

Welcome back everyone in this example we have at 25 degrees Celsius. What is the difference in entropy for our reaction entropy as well as gibbs free energy where we have one mole of potassium chlorate solid which decomposes into one mole of potassium chloride solid and two moles of oxygen gas. We need to determine whether this reaction will be spontaneous and if not well changing the temperature, make it spontaneous and if so should it be increased or decreased? So what we're going to begin with is verifying whether are given reaction is balanced and based on the amount of atoms on both sides of our equation given in the prompt we can verify that this is actually going to be a balanced reaction. So we don't need to worry about adding any coefficients because everything there means it's balanced right now. Our first step is to find our change in entropy for reaction. And so we should recall it to find the change in entropy. We're going to take the entropy of our products and we're gonna subtract that from the change in entropy of our reactant. And so we're gonna plug in our values and we're going to find these values from our prompt or sorry from our textbooks or online where for our product side we're going to begin off with our two moles of our oxygen gas which we will multiply by the standard molar entropy of oxygen gas which we can find in our textbooks or online and equals a value of 205.2 jewels divided by moles times kelvin then adding on to this because we have another product which is our one mole of our and sorry this should say two moles of 02. So we have our one mole of our potassium chloride. We're going to multiply this by potassium chloride standard molar entropy which we can find in our textbooks or online and is equal to a value of 82.6 jewels divided by moles times kelvin. So this completes the some of our entropy of our products. And now we're going to subtract this from the some of the standard entropy of our reactant. So beginning with our only reactant where we have one mole of potassium per chlorate, we're going to multiply potassium chlorate by its standard molar entropy we can find online or in our textbooks equaling the value 1 51 point oh jules divided by moles times kelvin. And so we should recognize that we'll be able to cancel our units of moles since we have in the numerator and the denominator in every parentheses here. So we're canceling that out and we're left with jewels per kelvin as our final unit. So plugging everything in our calculators, we're going to get that our change in entropy for reaction is equal to a value of 342 jewels per kelvin. Now we should recall that our change in entropy should be in units of kilo joules per kelvin. And so we're going to multiplied by the conversion factor where jules is in the denominator and kayla jules is our final unit in the numerator. And bergen to recall that our prefix kilo. So it tells us that we have 10 to the negative third power of our kilo joules equivalent to one Juul. And so this allows us to cancel out our unit of jewels leaving us with kilo joules per kelvin as our final unit. And this is going to yield an amount for the change in entropy for reaction equal to a value of 0.342 kg jewels per kelvin. And so this would be our first answer. Our change in entropy of our reaction. Next we want to find our change in entropy of our reaction so delta H. Of our reaction. And because this is another difference we're going to follow the same process where we take the change in standard entropy of our products. But in this case this is the standard entropy a formation of our products subtracted from the standard entropy a formation of our reactant. And so following this process here we're going to say that our change in entropy of our reaction is equal to first beginning with the difference in entropy of formation of our products. We're going to start off with our first product which again is our two moles of our oxygen which is multiplied by the entropy of formation of oxygen gas in its standard state because it's a di atomic molecule. That value should always be zero kg joules per mole. And again, that is due to the fact that this is oxygen standard state as a diatonic molecule. So you will see that when you look this up online or in your textbook you're going to get the value zero kg joules per mole. Now we're going to add this to our 2nd 2nd product where we have one mole of potassium chloride multiplied by potassium chloride entropy of formation which we can see in our textbooks or online is equal to the value negative 436.5 kg joules per mole. So this completes the some of our entropy of formation of our products. And now we're subtracting from the entropy of formation of our only reactant which is our one mole of our potassium per chlorate, Multiplied by potassium chlorate standard entropy of formation in our textbooks which we see is equal to the value -432.8. And we have units again of kilo joules per mole. So this completes the entropy of formation of our reactant and now plugging everything into our calculators. But also let's cancel our units first. So we can get rid of moles leaving us with kayla jules as our final unit. And we can say that the difference in entropy for the reaction is equal to a value of negative 3.7 kg joules. And so this would be our second answer here and now we have to determine our difference in Gibbs free energy for reaction. So what we're going to do is say that our difference and Gibbs free energy for reaction is equal to what we should recall as the difference between our entropy of our reaction subtracted from the temperature, multiplied by our change in entropy for reaction. And so going back to the prompt, we can see that we're given a temperature of 25°C. So recall that temperature should be in Kelvin for this formula. So we would take 25°C and add to 73.15 to give us 2 98. Kelvin. So we're gonna plug that in our formula below. And what we can say is that our change in entropy of our reaction which we determined above as negative 3.7 kg joules subtracted from our temperature which we can plug in as to 98.15 kelvin and then multiplied by our change in entropy of our reaction which above we determined As the value here where we had 0.342 so 0. killer jewels per Kelvin. We're going to first begin by taking the product of these two quantities here and that will give us negative 3.7 kg joules which is subtracted from the product which we should get of 101.9 67. And as far as our units will be able to cancel out kelvin with kelvin here, leaving us with kayla jewels. And so we have one on 1.9 67 kg jewels. And now taking this difference, we're going to get a value so we get a difference equal to the value negative 105.6673. And actually to be clear, we can cancel the unit of kilo jewels with kayla jewels here, meaning that at this step we did not need the key legal unit. And so we won't have units for our change in Gibbs free energy. We just have this value negative 105.6673. Now because this value is negative, we recognize that therefore our reaction is spontaneous and we can be more specific and say it's spontaneous at all temperatures. And so for our final answer, we can highlight our third final answer as our change in Gibbs Free energy here, calculated as this value here. And we're going to say that our reaction is therefore spontaneous because we have this negative Gibbs free energy value and the reaction will stay spontaneous at all temperatures. So everything highlighted in yellow represents our three final answers to complete this example. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

Verified Solution

Key Concepts

Thermodynamic Functions (ΔH, ΔS, ΔG)

Spontaneity of Reactions

Temperature's Effect on Spontaneity