Propane and Chlorine react in the presence of UV light to give 2-chloropropane and HCl. Estimate the enthalpy change of this reaction using the following bond enthaplies (KJ/mol) : C-H=+413, Cl-Cl=+243, C-Cl=+346 and H-Cl=+432.
- Write equation, ensuring both sides are balanced.CH3CH2CH3 + Cl2 → CH3CH(Cl)CH3 + HCl (2.) Draw structural formula of all reactants and products. (3.) Count the number of, and which types of bonds are present on both sides. Use the bond enthalpies in the question to total up both sides: Reactants: 8(C-H) = 8 * 413, 2(C-C), 1(Cl-Cl) = 243. Products: 7(C-H) = 7 * 413, 2(C-C), 1(C-Cl) = 346, 1(H-Cl) = 432. The bond enthalpy for (C-C) is not given in the question but we can cancel out it out because there are 2(C-C) on each side. Therefore; Reactants = 3547 KJ/mol and Products = 3669 KJ/mol (4.) Use formula Products - Reactants to get answer of +122KJ/mol. Remember the + sign!
JW
