estimate Delta H for the following reaction
C2H4 (g)+3O2(g)→2 CO2(g)+2 H2O(g)
So we have (4) C-H bonds and (1) C=C bond or (H = 4*414 + 615= 2271)
For oxygen,
(3) O=O bonds or (H = 3*502=1506)
2CO2 is (2) O=C=O
(H=730*4=2920)
2H2O is (2) H-O-H
or (H=4*464=1856)
Delta H =(2271+1506)-(2920+1586)=-729
But the actual answer is -1614.
Can you please tell me where am I wrong?
You are incorrect in your calculation of the enthalpy of formation of C2H4. The enthalpy of formation of C2H4 is actually +52 kcal/mol. This means that the enthalpy of the reaction is -1614 kcal/mol.
Your calculations for the enthalpy of formation of each molecule seem to be correct. However, in the calculation for the overall change in enthalpy (∆H) of the reaction, you made an error.
The balanced chemical equation for the reaction is:
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (g)
To calculate the ∆H for the reaction, you need to sum up the bond energies of the reactants and subtract the sum of the bond energies of the products.
Reactants:
C2H4: 4 C-H bonds (4 * 414 kJ/mol) + 1 C=C bond (615 kJ/mol)
3O2: 3 O=O bonds (3 * 498 kJ/mol)
Products:
2CO2: 2 O=C=O bonds (2 * 799 kJ/mol)
2H2O: 2 O-H bonds (2* 464 kJ/mol)
Now let's calculate the overall change in enthalpy (∆H):
∆H = (4 * 414 + 615) + (3 * 499) - (2 * 799 + 2 * 464)
= (1656 + 615) + (1497) - (1598 + 928)
= 2271 + 1497 - 2526
= -267
So, the correct estimate for the ∆H of the reaction is -267 kJ/mol, not -729 kJ/mol. There might have been an error in your calculation for the sum of the bond energies of the products.
In your calculation, you correctly identified the number of bonds present in each molecule and assigned a corresponding bond energy value. However, it seems that you made an error in your final calculation.
Let's go through the calculations again step by step:
1. Calculate the bond energy for the C-H bonds in C2H4:
(4 C-H bonds) * (414 kJ/mol) = 1656 kJ/mol
2. Calculate the bond energy for the O=O bonds in O2:
(3 O=O bonds) * (502 kJ/mol) = 1506 kJ/mol
3. Calculate the bond energy for the O=C=O bonds in CO2:
(2 O=C=O bonds) * (799 kJ/mol) = 1598 kJ/mol
4. Calculate the bond energy for the H-O-H bonds in H2O:
(2 H-O-H bonds) * (464 kJ/mol) = 928 kJ/mol
Now, let's calculate the overall enthalpy change (ΔH) for the reaction:
ΔH = (Σ bond energies of reactants) - (Σ bond energies of products)
= (1656 kJ/mol + 1506 kJ/mol) - (1598 kJ/mol + 928 kJ/mol)
= 3162 kJ/mol - 2526 kJ/mol
= 636 kJ/mol
Therefore, the estimated value for ΔH of this reaction is +636 kJ/mol (endothermic).
It seems that the actual answer you provided (-1614) is the opposite sign of the estimated value. This discrepancy might be caused by an incorrect assignment of signs or an error in calculations for one or more bond energies. Please double-check the bond energy values and the calculation steps to identify the source of the error.