When a gaseous compound X containing only C, H, and O
is burned i...

Question

When a gaseous compound X containing only C, H, and O
is burned in O2, 1 volume of the unknown gas reacts with
3 volumes of O2 to give 2 volumes of CO2 and 3 volumes
of gaseous H2O. Assume all volumes are measured at the
same temperature and pressure.
(d) Combustion of 5.000 g of X releases 144.2 kJ heat.
Look up ΔH°f values for CO21g2 and H2O1g2 in
Appendix B, and calculate ΔH°f for compound X.

Accepted Answer

Hi everyone. This problem reads compound A contains only carbon, hydrogen and oxygen in a reaction one volume of liquid A is burned in four volumes of oxygen gas to form three volumes of carbon dioxide and three volumes of gaseous H calculate the standard entropy of formation for A If the combustion of 3.65 g of A releases 100 and 12.4 kg joules of heat. So, our goal for this problem is to calculate the standard entropy of formation for A. So, from the problem, we can write out a balanced chemical equation. So we're told one volume of liquid A. So we're going to go ahead and write a and we have one mole of this is burned and four volumes of oxygen gas. Okay, so we're going to get our four volumes of oxygen gas to form three volumes of carbon dioxide. Okay, And three volumes of gaseous H 20. So, this is our balanced chemical equation. And from this we can see that we have three carbons, We have six hydrogen and we have one oxygen. So, what this means then is compound A is equal to C three H60. Okay, and what we want to do is we want to calculate the standard heat of this combustion or the standard entropy of this combustion and we know it's a combustion because in the problem, it tells us that it is burned. Okay, so in order for us to calculate the entropy of this combustion, it's going to equal kila jewels over moles. So, we know what the killer joules of heat is that value was given in the problem. The killer joules of heat is 100 and 12.4 kg joules of heat. So now we need to figure out our moles for compound A. And we can do this because we're told how many grams of compound A. We have, we're told in the problem that we have 3. g of compound A. So what we want to do here is we want to go from grams of a two moles of A. And we're going to do do this using the molar mass of compound A. So let's go ahead and write down the molar mass for compound A. And we'll do this using the periodic table. So the molar mass for C three H 60. Is 0.791 g per mole. So let's go ahead and convert that 3.65 g of a. Two moles of a. Alright, so let's go ahead. And right here this is 3.65 g of A. So we're going to go Are you were going to use the conversion and one mole of a. There is 58. 91 g of a. Alright. So here we'll see that grams of a cancel and we're left with moles of A. So once we do this we'll take 3.65 and divided by 58.791. And we get negative 1700 Oh, excuse me, we get zero, we get zero . moles of a. So let's go ahead and plug that in here. So for our moles of a we got 0. moles. So now we have kilograms over moles. So our entropy of combustion is going to be - 0.52 kg joules per mole. Alright, so now that we know our entropy of the combustion, we can write out our equation to calculate the standard entropy for a So the equation that we're going to use is the equation that we use to solve for the entropy of a reaction. Okay, so our entropy of the reaction or in this case of the combustion is going to equal the moles of the products, minus the reactant, multiplied by their standard heats of formation. Alright, so let's go ahead and write this out. So, for our products, we have carbon dioxide and water as a gas. So we're going to go ahead and write for our carbon dioxide, we have three moles of carbon dioxide. So we're going to take this three moles of carbon dioxide and multiply it by the standard heat of formation for carbon dioxide and this is going to be plus our second product is the H 20 as a gas and we have three moles of that. So it's going to be three times the standard entropy of formation for water as a gas. So this is the sum of our products minus the sum of our reactant reactant. We have the compound A. So we only have one mole of this. So we have one times the standard heat of formation for compound A. Which we wrote is C three H 60. Plus our second reactant, Let's make some room. So this is going to be plus our second reactant is oxygen gas. Alright, and we're gonna write this is going to be plus four times the standard heat of formation of oxygen gas. Alright, so this is our equation that we would use to solve for the standard the entropy of a reaction or for this case it's the combustion. So it's going to be the sum of our products minus the sum of our reactant. So what we want to solve for here is the heat of formation for A. So we want to solve for this. Okay, so we need to rearrange the equation so that we're solving for that. So what we're going to rearrange this to is the standard heat of entropy or a standard heat of formation for compound A is going to equal three times the standard entropy of formation for carbon dioxide plus three times the standard entropy of formation of water minus the entropy of reaction for the combustion. We're going to ignore the oxygen gas because the standard entropy of formation is going to be zero for that. So, what we need to do now here is right the values for our heats of formation. So let's go ahead and write that here. So these these are values we're going to look up. So the standard heat of formation four carbon dioxide is equal to negative 393.5 kg joules per mole. The standard heat of formation or H 20 gas is negative, 241.8 kg jewels Permal. Yeah. And the standard heat of formation four oxygen gas is zero killer jules, Permal, Which is why we ignored it up above. So now we have all the values that we need. So we're just going to go ahead and plug those values in. So, our standard heat of formation for compound A is equal to three times the standard kita formation for carbon dioxide gas. So negative 393.5 kg joules per mole plus three times the standard heat of formation for water as a gas. 200 negative 241.8 kg joules per mole. And this is going to be minus the entropy of the combustion, which we calculated above is negative. So we're going to do minus negative 1788. kg joules per mole. Okay, so once we do this calculation, we're going to get the standard entropy or heat of formation for compound A is equal to negative 117.38 kg joules per mole. And this is going to be our final answer. Okay, so that is the standard heat of formation or standard entropy of formation for A if the if the combustion of 3.65 g of a release is 112.4 kg joules of heat. That is it for this problem. I hope this was helpful.

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