The standard enthalpies of formation of gaseous propyne (C₃H₄), propylene (C₃H₆), and propane (C₃H₈) are +185.4, +20.4, and -103.8 kJ/mol, respectively. (b) Calculate the heat evolved on combustion of 1 kg of each substance.

Question

Accepted Answer

Welcome back everyone, ethyl alcohol and hydrogen are being put forward as cleaner alternatives to traditional fossil fuels, compare the efficiency of these fuels by calculating the amount of heat of combustion per kilogram of octane hydrogen and ethyl alcohol. So let's begin by writing out our combustion reaction of octane, recall that octane is represented by the formula C eight H 18. And in a combustion reaction, the liquid form of the fuel is going to react with oxygen gas and our products are always going to be C 02 and liquid water. So now we just need to make sure that this is balanced by adding the following coefficients will have an in front of water, a coefficient of 16 in front of carbon dioxide, a coefficient of 25 in front of oxygen and a coefficient of two in front of our octane and now we have a balanced reaction. So now we're going to recognize that since the prompt asks us for the heat of combustion, this is ultimately going to be able to be found by finding the entropy of our reaction for each of these fuels and converting to kill jules per kilogram. Because we would recall that entropy is reported in kilo jewels. So getting into that, we're going to have that our entropy of our reaction we want to recall is found by taking the entropy of our products minus the entropy of our reactant. So in the case of octane, we'll say that our entropy of our reaction is equal to the entropy Beginning with our products where we have 16 moles and let's use the color purple, we have 16 moles multiplied by the entropy of our first product, c. 0 2 Which is added to 18 moles multiplied by the entropy of water. And this completes the entropy of our products. Now we're going to subtract by the entropy of our reactant where we have two moles multiplied by the entropy of octane. So delta H of octane Added to 25 moles, multiplied by the entropy of oxygen. So going into our textbooks, we're going to reference our entropy values for each of our re agents. And beginning with our products will have 16 moles and let's actually keep the colors consistent. So we'll have 16 moles multiplied by the entropy of C. 02. Which we'll see in our textbooks is equal to a value of negative 93.5 kg joules per mole. And then added to this we have 18 moles multiplied by the entropy of water which in our textbooks will see as a value of negative 2 85.8 kg joules per mole. And so subtracting this ending off our back it's and subtracting this where we begin with the entropy of our reactant, we have two moles multiplied by the entropy of octane, which in our textbooks will see as a value of negative 208.4 kg jewels Permal. And then added to this, we have 25 moles multiplied by the entropy of oxygen. Which we want to recall is di atomic and exists as a gas here. Sorry, di atomic gas. So this is its standard state. So it's going to have an entropy of zero since its in its standard state. And this applies to any other molecule in its standard state. And so now that we have our entropy of our reactant filled in, we're just going to simplify this in our calculators. So that in our next line will have the values negative 1000 or sorry, negative 11,440 10000.4 kg joules because we'll see that our units of moles will cancel out subtracted from the value 416.8 kg joules once we cancel out moles. And so taking the difference here will find that our entropy of our reaction is equal to negative 11,857 0.2 kg jewels. And we're interpreting this as killed jules which um we're going to need to divide by two here since in our reaction we need the heat of combustion for one mole of octane. And in our reaction we have two moles of octane. So we're going to divide by two moles of our octane. And so in doing so this now gives us our entropy for one mole of octane equal to negative 5511.8 kg joules per mole of octane. And so now that we have this this entropy we just need this in kilograms per kilogram. So now we're just going to do a conversion factor to get to the heat of combustion for octane. Sorry, That's C eight H 18. And Killer joules per kilogram as the prompt asks. So beginning with what we just calculated, we have negative 5511.8 kg joules per mole of our octane. Which we want to multiply by a conversion factor to cancel out our units of moles in the numerator. And to do that we're going to use the molar mass of octane in grams per mole which from our Pyatykh cables will see that we have A value of 114.26 g of octane. And sorry, let's make that meter. So 114.26 g of octane for one mole of octane. So this allows us to now cancel out our moles of octane. And now we're left with killer jules program but we need kilograms per kilogram. So we're going to multiply by our next conversion factor to cancel out grams in the numerator and introduce kilograms in the denominator. Where will recall that our prefix kilo tells us that we have 10 to the third power of our base unit grams And canceling out grams were left with killer joules per kilogram as our final units for the heat of combustion. And this is going to result in a value of negative 48,239 killer joules per kilogram for our heat of combustion of octane. So this would be our first answer that we found for octane here. So let's move on to our next fuel mentioned in the prompt. So I'll just draw a line to break up the work. Our next fuel mentioned in the prompt is hydrogen. So we'll begin by writing out the combustion of hydrogen. So we have hydrogen gas which is a di atomic gas. So it exists as H2. It's going to react with oxygen gas in a combustion where as a product we're going to form just water since we don't have any carbon in hydrogen. So we have liquid water as our product. And now to make sure that this is balanced, we're going to place a coefficient of two in front of our water and a coefficient of two in front of our hydrogen gas. And now we have a bounced reaction where we can go right into calculating our entropy of our reaction here and so again, that is going to be the entropy of our products. So for our product, we have two moles of the entropy of water and then subtracting from this, we have our entropy of our reactant where we have two moles multiplied by the entropy of hydrogen gas added to our entropy of oxygen gas, since it's just one more and so plugging in our terms from our textbooks for the entire piece of our agents, we have first for our products to moles multiplied by the entropy of water, which in our textbooks we see is negative 2 85.8 kg joules per mole. This is then subtracted from the entropy of our reactant where we have two moles multiplied by the entropy of hydrogen gas, which again exists in its standard state as a di atomic gas. So it has an entropy of zero kg joules per mole. And then added to this, we have our entropy of oxygen gas which in our textbooks also exists as a diatonic molecule in its standard state. So it has an entropy of zero kg joules per mole as well. And so now we're just going to simplify this so that we find that our entropy of our reaction is now equal to a value of negative 571.6 kg jewels. But as we can see, we have in our equation two moles of hydrogen gas and that is what this entropy is for. So we're going to divide this by two moles of hydrogen gas and this is going to now give us an entropy that is equal to a value of negative 285.8 kg joules per mole of hydrogen gas. And so now we just need this in units of kilograms per kilogram to get our heat of combustion. So for a heat of combustion of hydrogen gas and jewels per kilogram, we would take our value, we just found negative 2 85.8 kg joules per mole of hydrogen gas. And multiply this to cancel out our units of moles. Where we introduce the molar mass of hydrogen gas, where in the numerator we have one mole of hydrogen gas. And in our periodic table we'll see that it has a mass of two point oh two g of H two. So canceling out moles of H two. We now want to get rid of grams by introducing kilograms. So we'll have grams in the numerator and kilograms in the denominator and recall again, our prefix kilo tells us we have 10 to the third power of our base unit grams, canceling out grams. We're left with killer joules per kilogram as our final units. And this results in a heat of combustion equal to 141, 485 kg jewels per kilogram. So this would be our second answer here that we found for the heat of combustion of our second fuel hydrogen gas. And so moving forward in our solution, we're going to do the same process for our third fuel mentioned in the prompt which was our ethyl alcohol. So for ethyl alcohol, we want to recall that that's another way of saying the compound ethanol and it has the formula C two H five oh H. So we're going to begin by writing out its combustion. So we have C two H five oh H reacting with oxygen gas and this is our fuel. So it's in liquid form. This is going to produce as a product carbon dioxide gas and liquid water. And so making sure that this is balanced, we're going to add a coefficient of three in front of water, a coefficient of two in front of carbon dioxide and a coefficient of three in front of oxygen for a balanced equation And now that we have our bounced equation, we're going to go ahead and calculate the entropy for this reaction. Where again we have the entropy of our product. So we'll do that in purple. We have our first product to moles multiplied by the entropy of C. 02 Added to three moles, multiplied by the entropy of water, completing the entropy of our products. Will subtracting this from the entropy of our reactant. Where we begin with our entropy of our ethanol Added to three moles, multiplied by the entropy of oxygen, completing our entropy of our reactant. And now plugging in our values from our textbooks for these re agents. We begin with our products where we have two moles multiplied by the entropy of carbon dioxide which in our textbooks we see as negative 93.5 kg joules per mole added to this. We have three moles multiplied by the entropy of water which we'll see in our textbooks is negative 2 85.8 kg joules per mole, completing the entropy of our products. And subtracting this from the entropy of our reactant. We have our entropy of ethanol which in our textbooks we see as the value of negative 77 point oh joules per mole. And adding to this we have three moles of our second reactant oxygen which has an entropy in our textbooks because it's in its standard state of zero kg joules per mole. And so now simplifying this in our calculators will find that our entropy of this reaction is equal to the difference between the entropy of our products which should be negative. 16,400 or sorry, 1,644.4 kg joules after we cancel out our units of moles. And then this is subtracted from our entropy of our reactant which should simplify to negative 2 77 point Okay illegals. So taking the difference here will find an entropy for the reaction equal to negative 1367. kg jewels. And since we already only have one mole of ethanol in our balanced reaction here we just have a coefficient of one. This is Permal of ethanol. And so now we're just going to get the heat of combustion as we did before by converting to kill jewels per kilogram for ethanol. And so we have our value. We just found negative 1367.4 kg joules per mole of ethanol multiplied by our first conversion factor to go from to go from grams to moles. But we want to cancel out moles. So we're going from moles to grams. So plugging in the molar mass of ethanol, we see on our periodic tables it has a value of 46.8 g of ethanol For one mole of ethanol. So canceling out moles. We're now going to introduce our unit kilograms by multiplying that next as a conversion factor. Where in the numerator we have 10 to the third power grams equivalent to one kg and getting rid of grams. That leaves us with our final units as killer joules per kilogram which is what we want. And this is going to simplify to a value equal to negative 29,674 kg or sorry kila jewels in the numerator, kila jewels per kilogram in our denominator as our third final answer for the heat of combustion of our third fuel ethanol. So everything highlighted in yellow represents our three final answers as our heat of combustion of each of our fuels and units of kill jules per kilogram for one mole of each of our gas is I hope 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

The standard enthalpies of formation of gaseous propyne (C₃H₄), propylene (C₃H₆), and propane (C₃H₈) are +185.4, +20.4, and -103.8 kJ/mol, respectively. (b) Calculate the heat evolved on combustion of 1 kg of each substance.

Verified Solution

Key Concepts

Standard Enthalpy of Formation

Combustion Reaction

Heat of Combustion

The standard enthalpies of formation of gaseous propyne (C3H4), propylene (C3H6), and propane (C3H8) are +185.4, +20.4, and -103.8 kJ/mol, respectively. (b) Calculate the heat evolved on combustion of 1 kg of each substance.

Verified Solution

Key Concepts

Standard Enthalpy of Formation

Combustion Reaction

Heat of Combustion

The standard enthalpies of formation of gaseous propyne (C₃H₄), propylene (C₃H₆), and propane (C₃H₈) are +185.4, +20.4, and -103.8 kJ/mol, respectively. (b) Calculate the heat evolved on combustion of 1 kg of each substance.