Question

Calculate the δh in kj/mol for the following reaction: c2h4(g) 3o2(g) → 2co2(g) 2h2o(g) bond enthalpies: c-h = 415 kj, c=c = 611kj , o=o = 498 kj , c=o = 741 kj , o-h = 464 kj

Answer 1

The enthalpy change (δH) for the reaction of ethylene with oxygen to produce carbon dioxide and water, given the bond enthalpies, is 1055 kJ.

Step-by-step explanation:

We need to calculate the enthalpy change (δH) for the combustion reaction of ethylene (C2H4) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). To find the heat of reaction, we must account for the energy required to break bonds in the reactants and the energy released when new bonds form in the products. Using the given bond enthalpies, we can calculate δH as follows:

• Breaking bonds in 1 mol of C2H4: 1 × C=C (611 kJ/mol) + 4 × C-H (415 kJ/mol each)
• Breaking bonds in 3 mol of O2: 3 × O=O (498 kJ/mol each)
• Forming bonds in 2 mol of CO2: 4 × C=O (741 kJ/mol each for 2 double bonds per CO2 molecule)
• Forming bonds in 2 mol of H2O: 4 × O-H (464 kJ/mol each)

The total energy absorbed in breaking the bonds is calculated by adding the individual energies:

1 × 611 kJ + 4 × 415 kJ + 3 × 498 kJ = 611 kJ + 1660 kJ + 1494 kJ = 3765 kJ

The total energy released in forming the new bonds is:

4 × 741 kJ + 4 × 464 kJ = 2964 kJ + 1856 kJ = 4820 kJ

Therefore, the net enthalpy change (δH) for the reaction is the energy released minus the energy absorbed:

δH = 4820 kJ - 3765 kJ = 1055 kJ