Topic 5.1 and 5.2 Hesss Law and Bond Enthalpies Hesss Law Topic 5.2 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) KJ shows three different pathways: A B A C B A D E B enthalpy change from reactants to products for all of these is the same if a series of reactions are added together, the enthalpy change for the net reaction ( H final ) will be the sum of the enthalpy change for the individual reactions ( H ind + H ind + H ind .) the change in enthalpy is the same whether the reaction takes place in one step, or in a series of steps H is independent of the reaction pathway depends only on the difference between the enthalpy of the products and the reactants H = H products H reactants provides a way to calculate enthalpy changes even when the reaction cannot be performed directly 8 energy in reactants energy in products Problem-Solving Strategy work backwards from the final reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal if a reaction is reversed the sign on H is reversed N 2 (g) + 2O 2 (g) 2NO 2 (g) H = 68kJ 2NO 2 (g) N 2 (g) + 2O 2 (g) H = - 68kJ multiply reactions by a coefficient to give the correct numbers of reactants and products in order to get the final reaction the value of H is also multiplied by the same integer identical substances found on both sides of the summed equation cancel each other out Example 1 Given: N 2 (g) + O 2 (g) 2 NO (g) H 1 = +181 kJ 2 NO (g) + O 2 (g) 2 NO 2 (g) H 2 = -113 kJ Find the enthalpy change for: N 2 (g) + 2 O 2 (g) 2 NO 2 (g ) H = H 1 + H 2 = +181 kJ + (-113 kJ) = + 68 kJ Example 2 H = (- 184 kJ) + ( kJ) = kJ Example 3 Given: C (s) + O 2 (g) CO 2 (g) H 1 = kJ mol -1 H 2 (g) + O 2 (g) H 2 O (g) H 2 = kJ mol -1 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) H 3 = kJ mol -1 Find the enthalpy change for: C (s) + 2H 2 (g) CH 4 (g ) This equation needs to be flipped. The CH 4 is on the wrong side of the equation Example 3 Given: C (s) + O 2 (g) CO 2 (g) H 1 = kJ mol -1 H 2 (g) + O 2 (g) H 2 O (g) H 2 = kJ mol -1 CO 2 (g) + 2H 2 O (g) CH 4 (g) + 2O 2 (g) H 3 = kJ mol -1 Find the enthalpy change for: C (s) + 2H 2 (g) CH 4 (g ) (- 393 kJ mol -1 ) + (- 572 kJ mol -1 ) + (+ 890 kJ mol -1 ) = - 75 kJ mol This reaction need to run twice/doubled because of the 2H 2 in the final equation Using enthalpy cycles instead counter-clockwise reaction energy (-3,222) has to equal clockwise reaction energy (-3,267 + H) H = +45 KJ/mol ? clockwise needs to equal counter-clockwise -109 = (-92.3) + H H = kJ mol -1 ? can be used to calculate the enthalpy change for a chemical reaction if we know the energy necessary to break or form bonds in the gaseous state breaking bonds energy is required so enthalpy is positive (endothermic) the molecule was stable so energy was necessary to break apart the molecule forming bonds energy is released so enthalpy is negative (exothermic) the new molecule is more stable than the individual atoms (unstable) that make it up, so energy is released Bond Enthalpies a molecule with strong chemical bonds generally has less tendency to undergo chemical change than does one with weak bonds SiO bonds are among the strongest ones that silicon forms it is not surprising that SiO 2 and other substances containing SiO bonds (silicates) are so common it is estimated that over 90 percent of Earth's crust is composed of SiO 2 and silicates we use average bond enthalpies again, in the gaseous state different amount of energy can be required to break the same bond example- methane, CH 4 if you took methane to pieces, one hydrogen at a time, it needs a different amount of energy to break each of the four C-H bonds every time you break a hydrogen off the carbon, the environment (think bond angles) of those left behind changes, and the strength of the remaining bonds is affected therefore, the 412 kJ mol -1 needed to break C-H is an average value and therefore the one used in calcuations The average bond enthalpies for several types of chemical bonds are shown in the table below: 20 Bonds broken 1 N N = 945 kJ 3 H-H 3(435) = 1305 kJ Total = 2250 kJ Bonds formed 2x3 = 6 N-H: 6 (390) = kJ Net enthalpy change = (+ 2250) + (- 2340) = - 90 kJ (exothermic) Calculate the enthalpy change for the reaction. Is it endo or exothermic? N H 2 2 NH 3 Bond Enthalpy Calculations Example 1: H-H Example 2 energy course of reaction 2H 2 + O 2 2H 2 O Working out H Show all the bonds in the reactants energy course of reaction 2H 2 O HH O=O+ Working out H Show all the bonds in the products energy course of reaction HH O=O + HH O HH O Working out H Show the bond energies for all the bonds energy course of reaction 436 O=O+ HH O HH O Working out H Show the bond energies for all the bonds energy course of reaction HH O HH O Working out H Show the bond energies for all the bonds energy course of reaction HH O 464 + Working out H Show the bond energies for all the bonds energy course of reaction Working out H Add the reactants bond energies together energy course of reaction Working out H Add the products bond energies together energy course of reaction Notice the negative due to bonds being formed Working out H H = energy in energy out energy course of reaction Working out H H = energy in energy out energy course of reaction Working out H H = energy in energy out energy course of reaction H = -486 exothermic