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Bond enthalpies question-help pls

https://www.quora.com/profile/Bravewarrior/p-146571273
Here is the question and its solution. I keep on getting +1262 as my answer. Can someone please explain why it should be -1262? Thanks!
(1) Quora - A place to share knowledge and better understand the world
I am stuck on this one too 😭
Any help would be greatly appreciated!
Reply 2
The formation of I2O5 & formation of 5CO + the reaction of I2O5 + 5CO = The formation of 5CO2 from hess cycle

Formation of I2O5 (-158) + the formation of 5CO (-550) is -708.
Formation of 5CO2 is -1970
To find enthalpy of reaction it's (formation of 5CO2) - (formation of I2O5 + formation of 5CO) if you rearrange
-1970 - (-708) = -1262 kJmol-1
Reply 3
Original post by pigeonwarrior
(1) Quora - A place to share knowledge and better understand the world
I am stuck on this one too 😭
Any help would be greatly appreciated!

Bonds in reactants:
There is one N=N (I cant type triple bond just imagine it is one)
3 x H-H bonds

so enthalpy for reactants is (3 x 436) + 941 = 2249

Reactants - products = enthalpy change
enthalpy for prods is 6 x N-H bonds (3 N-H bonds in each ammonia, and 2 molecules makes 6 bonds)
Need to find N-H bond enthalpy

2249 - 6N-H = -97
2249 + 97 = 6N-H
2346 = 6N-H
so 1 N-H bond enthalpy = 391 kJmol-1
Original post by hpxna
The formation of I2O5 & formation of 5CO + the reaction of I2O5 + 5CO = The formation of 5CO2 from hess cycle

Formation of I2O5 (-158) + the formation of 5CO (-550) is -708.
Formation of 5CO2 is -1970
To find enthalpy of reaction it's (formation of 5CO2) - (formation of I2O5 + formation of 5CO) if you rearrange
-1970 - (-708) = -1262 kJmol-1

Thank you so much! I was just confused as they used delta r h which means enthalpy change of reaction? No clue why they put that there then 😭 but thanks!
Original post by hpxna
Bonds in reactants:
There is one N=N (I cant type triple bond just imagine it is one)
3 x H-H bonds

so enthalpy for reactants is (3 x 436) + 941 = 2249

Reactants - products = enthalpy change
enthalpy for prods is 6 x N-H bonds (3 N-H bonds in each ammonia, and 2 molecules makes 6 bonds)
Need to find N-H bond enthalpy

2249 - 6N-H = -97
2249 + 97 = 6N-H
2346 = 6N-H
so 1 N-H bond enthalpy = 391 kJmol-1

Thank you for helping! 🙂
Reply 6
Original post by pigeonwarrior
https://www.quora.com/profile/Bravewarrior/p-146571273
Here is the question and its solution. I keep on getting +1262 as my answer. Can someone please explain why it should be -1262? Thanks!

probably the way u are putting it in the calculator
always use brackets!

should look like this
-1970-(-708)
sum of products - sum of reactants

plug this in and se what u get!
Original post by sisterr
probably the way u are putting it in the calculator
always use brackets!

should look like this
-1970-(-708)
sum of products - sum of reactants

plug this in and se what u get!

Yup, that might be the case. Thank you!! 🙂
Reply 8
Hi, you can only use bond enthalpies for reactions involving gases, you need to draw a Hess's law cycle for this reaction, if you need some help with Hess's law try https://www.science-revision.co.uk/A-Level_Hess's_law.html
Original post by Scanjo63
Hi, you can only use bond enthalpies for reactions involving gases, you need to draw a Hess's law cycle for this reaction, if you need some help with Hess's law try https://www.science-revision.co.uk/A-Level_Hess's_law.html

Thank youuu for pointing this out! It makes much more sense now 🙂

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