# Chemistry question- energetics

Hi, please could i have help on this question? I dont understand why the answer isn’t b?
Here is the question: https://i.ibb.co/CsDFdFk/IMG-0214.jpg
Thank you!
Original post by anonymous294
Hi, please could i have help on this question? I dont understand why the answer isn’t b?
Here is the question: https://i.ibb.co/CsDFdFk/IMG-0214.jpg
Thank you!

Think about ways you have been taught to calculate enthalpy changes at A level. If you try to name one equation in which enthalpy changes and activation energies are linked that you will have studied - I can say with certainty that nothing will come to mind.

So what other information can you use to work out enthalpy changes?
Original post by TypicalNerd
Think about ways you have been taught to calculate enthalpy changes at A level. If you try to name one equation in which enthalpy changes and activation energies are linked that you will have studied - I can say with certainty that nothing will come to mind.

So what other information can you use to work out enthalpy changes?

Would it be c because it can’t be a since electronegativity isn’t got to do with enthalpy changes, it can’t be b because again activation energy doesn’t explain the difference and there is no equation linking the two, it can’t be d becuase enthalpy of formation of i2 and cl2 is 0?
Original post by anonymous294
Would it be c because it can’t be a since electronegativity isn’t got to do with enthalpy changes, it can’t be b because again activation energy doesn’t explain the difference and there is no equation linking the two, it can’t be d becuase enthalpy of formation of i2 and cl2 is 0?

All of your logic is indeed correct- you can use bond enthalpies to calculate enthalpy changes of formation using Hess cycles.

It is also worth thinking about whether the statement is likely true or not without simply considering a process of elimination. Bond enthalpies tend to be smaller for longer bonds because the electrostatic attraction in the bond is weaker (by Coulomb’s law) and in this case, the orbital overlap in the H-I bond is much less effective. As such, the statement is true.
Original post by undefined
All of your logic is indeed correct- you can use bond enthalpies to calculate enthalpy changes of formation using Hess cycles.

It is also worth thinking about whether the statement is likely true or not without simply considering a process of elimination. Bond enthalpies tend to be smaller for longer bonds because the electrostatic attraction in the bond is weaker (by Coulomb’s law) and in this case, the orbital overlap in the H-I bond is much less effective. As such, the statement is true.

Ah yes, that makes more sense, thanks!!