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.
is answer 103 KJ
answered below.
To estimate the enthalpy change (ΔH) of the reaction 2SF4 + O2 → 2OSF4 using average bond enthalpies, you need to consider the bonds broken and formed in the reaction.
First, let's calculate the total bond enthalpy of the bonds broken:
2(S-F) bonds broken = 2 * 327 kJ/mol = 654 kJ/mol
1(O=O) bond broken = 1 * 495 kJ/mol = 495 kJ/mol
Now, let's calculate the total bond enthalpy of the bonds formed:
2(S=O) bonds formed = 2 * 523 kJ/mol = 1046 kJ/mol
4(S-F) bonds formed = 4 * 327 kJ/mol = 1308 kJ/mol
Next, subtract the total bond enthalpies of the bonds formed from the total bond enthalpies of the bonds broken to find the enthalpy change (ΔH) of the reaction:
Total bond enthalpies of bonds broken - Total bond enthalpies of bonds formed
(654 kJ/mol + 495 kJ/mol) - (1046 kJ/mol + 1308 kJ/mol)
= (1149 kJ/mol) - (2354 kJ/mol)
= -1205 kJ/mol
Therefore, the estimated enthalpy change (ΔH) of the reaction is -1205 kJ/mol.
Please note that the answer is in kilojoules per mole (kJ/mol), not 103 kJ.
To estimate the enthalpy change (ΔH) of the reaction using average bond enthalpies, we need to calculate the energy required to break the bonds in the reactants and the energy released when new bonds are formed in the products.
The reactants are 2 SF4 (sulfur tetrafluoride) molecules and 1 O2 (oxygen) molecule, while the products are 2 OSF4 (oxygen tetrafluoride-sulfur) molecules.
Breaking the bonds:
In 2 SF4 molecules, there are a total of 2 x 4 = 8 S-F bonds. Given that the bond enthalpy of S-F is 327 kJ/mol, the energy required to break these bonds is 8 x 327 = 2616 kJ.
In 1 O2 molecule, there is a double bond, so we need to break 1 x 2 = 2 O=O bonds. Given that the bond enthalpy of O=O is 495 kJ/mol, the energy required to break these bonds is 2 x 495 = 990 kJ.
Forming the bonds:
In 2 OSF4 molecules, there are a total of 2 x 4 = 8 S=O bonds. Given that the bond enthalpy of S=O is 523 kJ/mol, the energy released when these bonds are formed is 8 x 523 = 4184 kJ.
Calculating the enthalpy change:
Now, we need to sum up the energy required to break the bonds and subtract the energy released when the new bonds are formed:
ΔH = (Energy required to break bonds) - (Energy released when bonds are formed)
= (2616 kJ + 990 kJ) - (4184 kJ)
= -578 kJ
The negative sign indicates that the reaction is exothermic, meaning it releases energy. Therefore, the estimated enthalpy change (ΔH) of the reaction is -578 kJ/mol.
Hence, the answer is -578 kJ.