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 There is another way to calculate the heat of reaction, using bond enthalpies.  Bond enthalpy refers to the amount of energy stored in the chemical.

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Presentation on theme: " There is another way to calculate the heat of reaction, using bond enthalpies.  Bond enthalpy refers to the amount of energy stored in the chemical."— Presentation transcript:

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2  There is another way to calculate the heat of reaction, using bond enthalpies.  Bond enthalpy refers to the amount of energy stored in the chemical bonds between any two atoms in a molecule.  Bond enthalpies have been experimentally determined and can be found in a Table of Bond Enthalpies. (see table)

3  The net change in energy during a chemical reaction is the difference between how much energy it takes to break chemical bonds and how much energy is released when bonds form.  So if we know how much energy is stored in these chemical bonds, we can calculate the change in energy for a reaction. The energy change will equal:

4  The amount of energy required to break the bonds of the reactant molecules — The amount of energy released when the bonds of the products form

5  It is important to realize that bonds are broken on the reactant side of the equation, and bonds are formed on the product side.  This will give us an equation to use that is very similar to our Hess's Law formula, but with one important different - when using Hess's Law, we subtract the reactants from the products.  But when using bond enthalpies, we will subtract the products from the reactants.

6  To solve bond enthalpy questions, you'll need to be able to draw the structural formulas of models, something you likely learned in Chemistry 20.  You'll need to know which atoms are bonded together, and if single, double, or triple bonds are involved. We'll keep our molecules pretty simple here.

7  Using bond enthalpies, provided in the table below, calculate the heat of reaction, Δ H, for: ½ H 2 (g) + ½ Cl 2 (g) → HCl(g) Given the following bond enthalpies:  H — H 436 kJ Cl — Cl 2 43 kJ H — Cl 433 kJ (from bond table)

8  Draw structural formulas for all molecules.  Chemical formula Structural Formula H — H H2 Cl — Cl Cl 2 H — Cl HCl  Using the balanced equation and the structural formulas, determine how much energy is required to break all of the bonds of the reactants, and how much energy is released when all of the product bonds form:

9  Reactants: bond # of bonds Bond Enthalpy Total Energy H-H1 X 436 Cl-Cl1 X243 bond # of bonds Bond Enthalp y Total Energy H-H1 X 433 Products

10  We still need to use the balancing coefficients from the balanced equation:  For example, 1 mole of H - H bonds requires 436 kJ but in our balanced equation only ½ mole of H - H bonds is involved, which will require only 218 kJ (436 × 0.5):  Δ H = Σ (bonds broken) – Σ (bonds formed) Δ H = [½(436) + ½(243)] – [(433)] Δ H = 339.5 – 433 = -93.5 kJ

11  Using the bond enthalpies provided, calculate the heat of reaction, Δ H, for: C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g) Given the following table of bond enthalpies:  C — H 413 kJ C = C 6 14 kJ C — C 348 kJ H — H 436 kJ

12  1. Begin by drawing the structural formulas for all molecules.  Chemical formula Structure Formula C 2 H 4 C 2 H 6 H 2 H — H

13  2. Determine how many bonds of each type are present in each molecule, and calculate the bond energies.  For example, in C 2 H 4 there is a single C=C bond (a double bond) but there are four C - H bonds.

14  Reactants Products Bond# of bonds Bond Energy Total Energy # of bonds Bond Energy Total Energy C - H 4 413 165264132478 C = C1614 H- H1436 C - C1348

15  3. Finally calculate Δ H for the reaction, check coefficients from the balanced equation:  C2H4(g) + H2(g) → C2H6(g)  Δ H = Σ (reactant bonds) – Σ (product bonds)  Δ H = (1652 + 614 + 436) – (2478 + 347)  Δ H = 2702 – 2826 = -124 kJ

16  1.2.6- assignment


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