4. A) What is the reaction specified for finding the heat of formation for C2H26

Question

4. A) What is the reaction specified for finding the heat of formation for C2H26 ? Don't forget states! Remember that the heat of formation is the energy absorbed or released when one mole of the substance is made from its elements in their standard states. B) From the following data: soi (g) + H20 (l) H2SO4 (1) 2CsHis (g) + 25 O2 (g) 16CO2 (g) +18H20 (1) C (graphite) + O2 (g) CO2 (g) C (graphite) + ½ O2 (g) CO (g) 2 C12H26 (g) + 3702 (g) 24 C0; (g) + 26 H2O (l) 2S(s) +3 O2 (g) 2 SO3 (g) H2 (g) + ½ 02 (g) H2O (l) AH- -88.09 kJ/mole AKPna=-11 078.4 kJ/mole AH 393.5 kJ/mole AH‰=-110.5 kJ/mole AH 15398.0 kJ/mole 11°nn-:-79 1.4 kJ/mole aHom=-285.8 kJ/mole (c) (f) Calculate the enthalpy of formation for C12H26. Be sure to SHOW ALL WORK! HINT! Use Hess's Law to rearrange SOME of the given reactions (b -h) to produce reaction a. Reaction a does NOT come from just rearranging the reactions given above! You will not need all reactions given, you should use three of them. Rewrite the required reactions here to get reaction a, and then find the enthalpy of formation, mathematically, for the compound of interest here: Reactions Enthalpies

Explanation / Answer

Enthalpy of formation reaction of C12H26 can be written as

12 C(grahite) + 13 H2 (g) C12H26(g)                Hf0 = ?

Using 3 equations we can solve this

C(graphite) + O2(g)         CO2(g)   Hrxn0 = -393.5 kJ/mol (d)

2 C12H26 (g) + 37 O2(g)     24 CO2(g) + 26 H2O(l)    Hrxn0 =  - 15398.0 kJ/mol (f)

H2(g) +   1/2 O2(g)         H2O(l) Hrxn0 = - 285.8 kJ/mol (h)

Multiply equation (d) with 12

12 C(graphite) + 12 O2(g)     12 CO2(g)   Hrxn0 = 12 X -393.5 kJ/mol = - 4722.0 kJ/mol.......(1)

Reversing equation (f) and divide by 2

12 CO2(g) + 13 H2O(l)     C12H26 (g) + 37/2 O2(g)   Hrxn0 = + (15398.0)/2 kJ/mol =+ 7699.0 kJ/mol.......(2)

Multiply equation (h) with 13

13 H2(g)   +   13/2 O2(g)     13 H2O(l) Hrxn0 = - 285.8 X13 kJ/mol = - 3715.4 kJ/mol........(3)

Adding equations (1),(2) and (3)

12 C(grap)+12 O2(g)+12 CO2(g) + 13 H2O(l)+13 H2(g)+13/2 O2(g) 12CO2(g)+C12H26(g)+37/2O2(g) +13 H2O(l)  

Cancelling like terms on both sides

12 C(grahite) + 13 H2 (g) C12H26(g)               

Hf0 = - 4722.0 kJ/mol ++ 7699.0 kJ/mol +- 3715.4 kJ/mol = - 738.4 kJ/mol

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