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Topic: Bond energy  (Read 202 times)

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Offline INeedSerotonin

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Bond energy
« on: February 24, 2020, 09:13:29 AM »


Hello

I was given this exercise that asked me to find the bond enthalpy of gaseous water. The left side says "Bond", and the right side says "Enthalpy".

I learnt that I have to calculate the enthalpy of the reagents minus the enthalpy of the products, right? So I did it.

2 H2 + O2 --> 2 H2O

436 x 2 + 498 - 4 x 464 = 486 kJ

However, if I divide all of the equation by two, I will have this:

436 + (498/2) - 2 x 464 = - 243 kJ, which is the correct answer.

Why does it change? Shouldn't it be the same? I cannot understand this. How am I to know the correct way to balance the equation then?

Thank you

Offline mjc123

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Re: Bond energy
« Reply #1 on: February 24, 2020, 11:37:36 AM »
Why does it change? Why not? Can you think of a number that doesn't change when you divide it by 2?
If the question really asked you for the "bond enthalpy" of water, the answer is 464 kJ/mol - the bond enthalpy of the O-H bond. Or 928 kJ/mol counting 2 OH bonds to a water molecule. This would be the enthalpy of the reaction
H2O(g)  :rarrow: 2H(g) + O(g)
What you have calculated is the enthalpy of formation of water. This is -243 kJ/mol. You got 486 (should be -486) the first time because that is the enthalpy of formation of 2 moles of water. It doesn't matter how you balance the equation, as long as you divide the answer by the coefficient of H2O in the equation. Do you understand why?

Offline INeedSerotonin

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Re: Bond energy
« Reply #2 on: February 24, 2020, 11:52:46 AM »
Why does it change? Why not? Can you think of a number that doesn't change when you divide it by 2?
If the question really asked you for the "bond enthalpy" of water, the answer is 464 kJ/mol - the bond enthalpy of the O-H bond. Or 928 kJ/mol counting 2 OH bonds to a water molecule. This would be the enthalpy of the reaction
H2O(g)  :rarrow: 2H(g) + O(g)
What you have calculated is the enthalpy of formation of water. This is -243 kJ/mol. You got 486 (should be -486) the first time because that is the enthalpy of formation of 2 moles of water. It doesn't matter how you balance the equation, as long as you divide the answer by the coefficient of H2O in the equation. Do you understand why?

Thank you! That's a very detailed answer. I understood everything.  ;)

Offline Enthalpy

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Re: Bond energy
« Reply #3 on: February 24, 2020, 12:12:27 PM »
Slightly beyond the question...

Bond energies are useful as an understanding tool. They are not accurate enough to compute reaction enthalpies. Bonds interfere, they can be deformed, and the heat of condensation is big. Use enthalpies of formation of complete compounds for that goal, measured (beware most data is estimated by software hence wrong) and in the desired state, liquid for instance.

Some substances have bonds that are not obvious, so take care not to forget them. When a reaction writes "C" or "Zn", it stands for a solid with many strong bonds, not for atomic carbon or zinc.

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