The enthalpy change ΔHo for 2 C (s) + 3 H2 (g) → C2H6 (g) is

Question

 Consider the following reactions:

 

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

 

               H₂ (g) + ½ O₂ (g) → H₂O (l)                      ΔH^o = − 286 kJ

 

               C (s) + O₂ (g) → CO₂ (g)                        ΔH^o = − 393 kJ

 

       The enthalpy change, ΔH^o, for 2 C (s) + 3 H₂ (g) → C₂H₆ (g) is

Answer 1

Hess' Law

2C(s) + 3H2(g) ==> C2H6(g) ... TARGET EQUATION

Given:

eq.1: C₂H₆ (g) + 7/2 O₂ (g) ==> 2 CO₂ (g) + 3 H₂O (l) ... ΔH^o = − 1541 kJ

eq.2: H₂ (g) + ½ O₂ (g) ==> H₂O (l) ...  ΔH^o = − 286 kJ

eq.3: C (s) + O₂ (g) ==>  CO₂ (g) ...   ΔH^o = − 393 kJ

Procedure:

reverse eq.1: 2CO2(g) + 3H2O(l) ==> C2H6(g) + 7/2O2(g) ... ∆Hº = +1541 kJ

copy eq.2 x3: 3H2(g) + 3/2O2(g) ==> 3H2O(l) ... ∆Hº = 3x-286 kJ = -858 kJ

copy eq.3x2: 2C(s) + 2O2(g) ==> 2CO2(g) ... ∆Hº = 2x-393 kJ = -786 kJ

Add these 3 equations and cancel and combine like terms....

2CO2(g) + 3H2O(l) + 3H2(g) + 3/2O2(g) + 2C(s) + 2O2(g) => C2H6(g) + 7/2O2(g) + 3H2O(l) + 2CO2(g)

2C(s) + 3H2(g) ==> C2H6(g) TARGET EQUATOIN

∑∆Hº values = 1541 kJ - 858 kJ - 786 kJ

∆Hº _rxn = -103 kJ

Answer 2

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 ΔH^o = 3×(− 286 kJ) = -858 kJ

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

Overall, -1644 kJ

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

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

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

Meanwhile, https://webbook.nist.gov/cgi/cbook.cgi?ID=C74840&Mask=2F gives a value of ΔH^o = − 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.