stilll didn't get it .. .. can u plz explain in detail ... how did u get the answer.. plz plz plz plz plzp l zp
thnks
Bond enthalpy is the energy required to break a mole of a certain type of bond.
O=O = 495 kj/mol
S-F = 327 kj/mol
S=O = 523 kj/mol
Use average bond enthalpies to estimate the enthalpy delta H (rxn) of the following reaction:
2SF4 + O2 ---- 2OSF4
Express your answer numerically in kilojoules.
Instead of asking me to work the problem step by step it would help if you told us what you don't understand. Not getting it doesn't tell us much. I gave you the formula.
Delta Hrxn = BEreactants - BEproducts =
[2*(4*S-F bond)]+ (O=O bond) - [2*(4*S-F bond) + (S=O bond)]= delta Hrxn
This is a matter of substituting those values for the bonds in the equation above to arrive at the answer. If you still don't understand, post your work showing how you arrived at the 103 value you posted earlier.
i am getting -28 KJ .. is it rite plz chk
That isn't what I obtained when I tried it. Show your work and let me see what you are doing wrong (or if I have made an error).
To estimate the enthalpy change (ΔH) of the reaction using average bond enthalpies, you need to consider the bonds broken and formed in the reaction and calculate the difference between the bond energies.
Given that the bond enthalpies are:
O=O = 495 kJ/mol
S-F = 327 kJ/mol
S=O = 523 kJ/mol
Let's analyze the reaction:
2SF4 + O2 → 2OSF4
In this reaction, we have bonds being broken in SF4 and O2, and bonds being formed in OSF4.
First, calculate the energy required to break the bonds. The bonds being broken are:
4 S-F bonds in 2SF4
1 O=O bond in O2
Energy required to break S-F bonds:
4 S-F bonds × 327 kJ/mol = 1308 kJ/mol
Energy required to break O=O bonds:
1 O=O bond × 495 kJ/mol = 495 kJ/mol
Next, calculate the energy released when new bonds are formed. The bonds being formed are:
4 O-S bonds in 2OSF4
8 S-F bonds in 2OSF4
Energy released when O-S bonds are formed:
4 O-S bonds × 523 kJ/mol = 2092 kJ/mol
Energy released when S-F bonds are formed:
8 S-F bonds × 327 kJ/mol = 2616 kJ/mol
Now, calculate the overall energy change:
Total energy required to break bonds = Energy required to break S-F bonds + Energy required to break O=O bond
Total energy required to break bonds = 1308 kJ/mol + 495 kJ/mol = 1803 kJ/mol
Total energy released when bonds are formed = Energy released when O-S bonds are formed + Energy released when S-F bonds are formed
Total energy released when bonds are formed = 2092 kJ/mol + 2616 kJ/mol = 4708 kJ/mol
Finally, calculate the ΔH (enthalpy change) of the reaction using the equation:
ΔH = Energy released - Energy required
ΔH = Total energy released when bonds are formed - Total energy required to break bonds
ΔH = 4708 kJ/mol - 1803 kJ/mol
ΔH = 2905 kJ/mol
Therefore, the estimated enthalpy change (ΔH) of the reaction 2SF4 + O2 → 2OSF4 is 2905 kJ/mol.