The Enthalpy Change ΔHo For 2 C (s) + 3 H2 (g) → C2H6 (g) Is

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This is a Hess' Law problem.

By definition (elements in standard state) C (s), H2(g), O2(g) all start at H = 0.

3 H2(g) + 3/2 O2(g) → 3H2O(l);  net ΔHo = 3×(− 286 kJ) = -858 kJ

2C(s) + 2O2(g) → 2CO2(g); net ΔHo = 2×(− 393 kJ) = -786 kJ

Overall, -1644 kJ

C2H6 (g) + 7/2 O2 (g) → 2 CO2 (g) + 3 H2O (l)      ΔHo = − 1541 kJ

so, C2H6 (g) + 7/2 O2 (g) is + 1541 from -1644, or:

2 C(s) + 3 H2 →  C2H6 (g) is ΔHo = − 103 kJ

Meanwhile, https://webbook.nist.gov/cgi/cbook.cgi?ID=C74840&Mask=2F gives a value of ΔHo = − 84 kJ, which is reasonably good agreement.

To help you understand this draw a diagram showing the relative energy levels of the various compositions.

  1. The elements (all at zero)
  2. The combustion products (2 CO2 (g) + 3 H2O (l)) (-1644 kJ)
  3. Ethane + oxygen (up by 1541 kJ from the combustion products).

To be sure you understand this, do the same exercise for the heat of formation of ethylene (C2H4) from the elements. Here are the steps:

  1. Write a balanced equation for the complete combustion of ethylene; note that the standard enthalpy of combustion for ethylene is -1411 kJ/mol (https://webbook.nist.gov/cgi/cbook.cgi?ID=C74851&Mask=1)
  2. Find the heat of formation for the combustion products, using the data given in this problem.
  3. Create the Hess' law diagram. As a check, the standard enthalpy of formation for ehtylene is +53 kJ/mole.

Từ khóa » C2h6 + 7/2o2 → 2co2 + 3h2o