Here is what I mean: take a carbon-carbon single bond (C−C). C−O (360); C=O (799); O=O (498); O−H (464), ΔH = [(3) (498) + (6) (414) + (2) (360)] − [(4) (799) + (6) (464)]. and then: ({+413x6} {+612} + {7/2 x +497} ) - ( {6x+463} + {2CO2}) which should give you the rough enthalpy of combustion of C2H6. Here is where I got most (not all) of the bond enthalpy values used in the problems below. Bonus Example #2: Given that a chlorine-oxygen bond in ClO2(g) has an enthalpy of 243 kJ/mol, an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the standard enthalpy of formation of ClO2(g) is 102.5 kJ/mol, use Hess's law to calculate the value for the enthalpy of formation per mole of ClO(g). The combustion of ethane proceeds in the gas phase according to the following equation, C2H6 (g) + 7/2 O2 (g) → 2 CO2 (g) + 3 H2O (g), (a) Estimate the enthalpy of combustion of C2H6 using the bond enthalpies, to find out the enthalpy combustion of ethane, use the bond enthalpies youve been given to work out the bonds formed + broken. 3) Adjust the given reactions as follows: 5) Add the three equations and their enthalpies to get: 6) That is the enthalpy to disrupt four Si-F bonds, so divide by 4 to get: which, to three significant figures, rounds off to +597 kJ. Note the use of 2x because there are two ClF molecules. Example #4: Calculate the bond energy of the Cl−F bond using the following data: Bond enthalpies (in kJ/mol): Cl−Cl (239); F−F (159). Express your answers in kJ/mol and kJ/g. 2) You get energy out when a bond (any bond) forms. For each product, you multiply its ΔH_"f"^° by its coefficient in the balanced equation and add them together. Calculate enthalpy of formation of methane (CH4) from the following data : (i) C (s) + O2 (g)→ CO2 (g), ΔrH° = – 393.5 kJ mol–1, (ii) H2 (g) +1/2 O2 (g) →H2O(l),ΔrH° = – 285.8 kJ mol–1, (iii) CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (l),ΔrH° = –890.3 kJ mol–1. The bond enthalpy for that situation would be different if six chlorines were instead attached to the carbons. Notice that, chemically speaking, H3O+ and H+ stand for the same thing. Also, a reminder: Now, I want to rearrange the negative sign on the second value in step three above. In this reaction, you are breaking 12 C-H bonds, 2 C-C bonds, and 7 O=O bonds. Example #1: Hydrogenation of double and triple bonds is an important industrial process. We’ve been given the standard enthalpies of formation of methane, carbon dioxide, and water. (i) and then subtracting eqn. The enthalpy change of this reaction will be equal to the standard enthalpy of combustion of methane. 1) The equation, as given, is not ready to be examined using Hess' Law. Comment: you may have noticed that four C−H bonds were involved on the reactant side and three C−H bonds were involved on the product side. Example #4 has a little trick in it. Do not forget that C−C bond! Standard Enthalpy of Combustion of Methane. Eliminate two H+ for the equation: 3) One last step to get the equation ready. Classification of elements and periodicity in properties, General principles and process of Isolation of metals, S - block elements - alkali and alkaline earth metals, Purification and characteristics of organic compounds, Some basic principles of organic chemistry, Principles related to practical chemistry. Calculate (in kJ) the standard enthalpy change ΔH for the hydrogenation of ethyne (acetylene) to ethane: Bond enthalpies (in kJ/mol): C−C (347); C≡C (839); C−H (413); H−H (432). Still have questions? Calculate enthalpy of formation of methane (CH. In this section we will use Hess's law to use combustion data to calculate the enthalpy of reaction for a reaction we never measured. Let's break all the bonds of the reactants: Note there are two C−H bonds in one molecule of C2H2 and there is one H−H bond in each of two H2 molecules. The ΔH is given per mole of HF, so we need to use −269 x 2 = −538 kJ for the enthalpy of the reaction. 4) Now, use a second bond enthaphy calculation for the equation in step 1 above: A different approach to solving this problem may be found here. Get your answers by asking now. Example #6: Calculate the enthalpy change for the reaction of ethene and hydrogen, given the following bond energy values in kJ/mol: List the bonds broken and the bonds made: Note that four C−H bonds were removed from each side. One final note before solving some problems: the ΔH values determined via this technique are only approximations. Calculate the enthalpy of formation of methane, given that the enthalpies of combustion of methane, asked Oct 31, 2019 in Chemical thermodynamics by Saijal (65.5k points) chemical thermodynamics; 0 votes.


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