Science, Tech, Math › Science Use Bond Energies to Find Enthalpy Change Determining the Change in Enthalpy of a Reaction Share Flipboard Email Print Enthalpy is the energy of a system. PM Images / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. our editorial process Facebook Facebook Twitter Twitter Anne Marie Helmenstine, Ph.D. Updated December 09, 2019 You can use bond energies to find the enthalpy change of a chemical reaction. This example problem shows what to do. Review You may wish to review the laws of thermochemistry and endothermic and exothermic reactions before you begin. A table of single bond energies is available to help you. Enthalpy Change Problem Estimate the change in enthalpy, ΔH, for the following reaction: H2 (g) + Cl2 (g) → 2 HCl (g) Solution To work this problem, think of the reaction in terms of simple steps: Step 1 The reactant molecules, H2 and Cl2, break down into their atoms. H2(g) → 2 H(g)Cl2(g) → 2 Cl(g) Step 2 These atoms combine to form HCl molecules. 2 H (g) + 2 Cl (g) → 2 HCl (g) In the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction: ΔH1 = +(436 kJ + 243 kJ) = +679 kJ Bond breaking requires energy, so we expect the value for ΔH to be positive for this step. In the second step of the reaction, two moles of H-Cl bonds are formed. Bond breaking liberates energy, so we expect the ΔH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ: ΔH2 = -2(431 kJ) = -862 kJ By applying Hess's Law, ΔH = ΔH1 + ΔH2 ΔH = +679 kJ - 862 kJΔH = -183 kJ Answer The enthalpy change for the reaction will be ΔH = -183 kJ.