I am beyond desperate. I do not understand how to calculate heat change at all and I have been looking online and reading my book for the last three hours. Below is posted a problem. It would be amazing if someone could work through it and explain each step they are doing and why they are doing it.
Calculate the heat change for the formation of lead(IV) chloride by the reaction of lead(II) chloride with chlorine
PbCl2(s) + Cl2(g) -> PbCl4(l) H=?
___
Use the following thermochemical equations:
Pb(s) + 2Cl2(g) -> PbCl4(l) h= -329.2kj
Pb(s) + Cl2(g) -> PbCl2(s) h= -359.4kj
Hey! I know this post is from like 13 years ago but in case anyone else comes across something like this:
Basically, we're trying to recreate the eq'n PbCl2(s) + Cl2(g) -> PbCl4(l) from the given 2 eq'ns, bc if we add them up, we'll get the overall original eq'n.
NOTE:
Hess's law says that no matter the path that a rxn takes, the overall enthalpy will be the same; here we'll add up the enthalpies of each individual eqn which will give us a final overall enthalpy which is what we're trying to find
Before we do all that, we need to recreate the OG eq'n as mentioned before, to do this, we have to change around the given 2 eq'n's first
1) to start, figure out which substances are needed on each side; the OG eq'n has PbCl2 on the reactants side and our second given eq'n is the one with PbCl2 - BUT the PbCl2 is on the products side which is weird!
but its all good bc we can switch the entire thing to get it on the reactants side
NOTE: when we flip it over, we also need to change the sign of H to positive
BEFORE:
Pb(s) + Cl2(g) -> PbCl2(s) H = -359.4 Kj
AFTER:
PbCl2(s) -> Pb(s) + Cl2(g) H = 359.4 Kj
ok cool we did it ... now that we have that sorted out, we need to see if the first eq'n needs any modifications:
Pb(s) + 2Cl2(g) -> PbCl4(l) H = -329.2 Kj
everything seems to be ok since we need PbCl4 on the products side and its already there so we can j leave it alone
now we have to add up the eq'n's to make the OG,
PbCl2(s) -> Pb(s) + Cl2(g) H = 359.4 Kj
Pb(s) + 2Cl2(g) -> PbCl4(l) H = -329.2 Kj
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start by cancelling out any double stuff - if u see the same substance on both sides cross it out on both sides:
here, Pb is a double so take that away from both sides
PbCl2(s) -> Cl2(g) H = 359.4 Kj
2Cl2(g) -> PbCl4(l) H = -329.2 Kj
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there's also Cl2 on both sides, but the products side of eq'n 2 has 1 extra, we can only take away 1 Cl2 from each side bc the reactants side of eqn 1 only has 1, so j take away 1 from each side
PbCl2(s) -> H = 359.4 Kj
1Cl2(g) -> PbCl4(l) H = -329.2 Kj
-------------------------------------
now put together whatever is left
PbCl2 + Cl2 -> PbCl4
WOAH! thats the OG eq'n omg were geniuses
so back to hess's law - to find the final enthalpy, we j add up the individual H from each eq'n and we'll get our final answer :
H = 359.4 Kj + (-329.2 Kj) = 30.2 Kj and there's your answer!
I flipped the second equation around to go with Hess's law which gives me a h value of 359.4 for the second one but I am lost after that.
I think you're trying to make it too hard.
Use equation 1 as is, reverse equation 2 (and change the sign of delta H for equation 2), add 1 to 2 and add delta H1 to the new delta H 2.
Pb(s) + 2Cl2(g) ==> PbCl4(l)
PbCl2(s) ==>Pb(s) + Cl2(g)
-----------------------------
Pb(s)+2Cl2(g)+PbCl2(s)==>PbCl4(l)+Pb(s)+Cl2(g)
Note that Pb(s) cancels.
Note that 2Cl2 on the left and 1 Cl2 on the right leaves 1 Cl2 on the left to leave you with
Pb(s) + Cl2(g) ==>PbCl4(l)
-329.2 + 359.4 = ??
to Drbob222 you actually are left with PbCl2(s) + Cl2(g) ==> PbCl4 because theres 2Cl2 on the first equation and the reverse equation only cancels out one of those Cl2's
Sure! I'll help you work through the problem step by step and explain each step along the way.
To calculate the heat change for the formation of lead(IV) chloride by the reaction of lead(II) chloride with chlorine, we need to use the given thermochemical equations and apply them in a way that allows us to cancel out the common reactants and products.
Let's start by writing the given thermochemical equations:
1. Pb(s) + 2Cl2(g) -> PbCl4(l) ΔH = -329.2 kJ
2. Pb(s) + Cl2(g) -> PbCl2(s) ΔH = -359.4 kJ
The reaction we're interested in is the formation of lead(IV) chloride, which involves the reaction of lead(II) chloride (PbCl2) with chlorine (Cl2). However, we don't have a direct equation for the formation of PbCl4. So, we need to rearrange and combine the given equations in a way that allows us to cancel out the common reactants and products.
We can see that in equation 1, one Pb atom and two Cl2 molecules react to form one PbCl4 molecule. In equation 2, one Pb atom and one Cl2 molecule react to form one PbCl2 molecule.
To obtain the desired overall reaction, we need to cancel out the common species, which are Pb(s) and Cl2(g). We can do this by adding equation 1 and equation 2 together:
Pb(s) + 2Cl2(g) + Pb(s) + Cl2(g) -> PbCl4(l) + PbCl2(s)
When we add the two equations together, we get:
2Pb(s) + 3Cl2(g) -> PbCl4(l) + PbCl2(s)
Now, let's calculate the overall heat change (ΔH) for this reaction.
Since we added the two equations together, we also need to add their respective ΔH values:
ΔH = -329.2 kJ + (-359.4 kJ)
ΔH = -688.6 kJ
The overall heat change for the formation of lead(IV) chloride is -688.6 kJ.
I hope this explanation helps you understand the process of calculating heat change and how to apply thermochemical equations to solve problems like this. Let me know if you have any further questions!