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MOLAR HEAT OF NEUTRALIZATION 300 ml of 0.2M aqueous KOH neutralizes 150ml of aqueous 0.2M H2SO4. The temperature rises from 22.3oC to 29.2oC. Calculate the molar heat of neutralization of KOH. (Assume the specific heat capacity of KOH and H2SO4 is the same as water)

How to solve these types of problems! 300 ml of 0.2M aqueous KOH neutralizes 150ml of aqueous 0.2M H2SO4. The temperature rises from 22.3oC to 29.2oC. Calculate the molar heat of neutralization of KOH. DH = -Q / n 1) Q=m x c x Dt =(300+150)*4.19*(29.2-22.3) = 13009.95 J = 13.01 kJ Now we must find the # of moles.

300 ml of 0. 2M aqueous KOH neutralizes 150ml of aqueous 0. 2M H2SO4 300 ml of 0.2M aqueous KOH neutralizes 150ml of aqueous 0.2M H2SO4. The temperature rises from 22.3oC to 29.2oC. Calculate the molar heat of neutralization of KOH. Molarity = n v n=molarity * volume n = 0.2 (0.3 L) = 0.06 moles of KOH Now we must calculate DH

300 ml of 0. 2M aqueous KOH neutralizes 150ml of aqueous 0. 2M H2SO4 300 ml of 0.2M aqueous KOH neutralizes 150ml of aqueous 0.2M H2SO4. The temperature rises from 22.3oC to 29.2oC. Calculate the molar heat of neutralization of KOH. DH = -Q / n = -13.01 / 0.06 = -216.83 kJ/mol

Hess' Law

Calculating D H: Hess’ Law The third way to calculate DH is to use a technique developed by Germain Henri Hess. Hess' Law states that the enthalpy of a reaction is independent of whether the reaction occurs in one or several steps.

Using Hess’ Law we can algebraically add the given equations and their accompanying DH's to obtain the DH for the desired or target equation.

Keep the following rules in mind 1) If an equation is multiplied or divided by a number, that factor also applies to DH. 2) If an equation is reversed, then the sign of DH changes.

3) Remember that enthalpy changes with different states of matter 3) Remember that enthalpy changes with different states of matter. Do not, for example, interchange H2O(l) with H2O(g). .

EXAMPLE #1 (do not copy!) Carbon disulfide is a very flammable solvent. It burns according to the following equation: CS2(l) + 3 O2(g) --> CO2(g) + 2 SO2(g)

CS2(l) + 3 O2(g) --> CO2(g) + 2 SO2(g) Calculate DH for the above reaction using the following data: (1) C(s) + 2 S(s) --> CS2(l) DH = 88 kJ (2) C(s) + O2(g) --> CO2(g) DH = -394 kJ (3) S(s) + O2(g) --> SO2(g) DH = -297 kJ

How to solve this problem?? 1) First look at the final equation. Look at the reactants and look at the products. CS2(l) + 3 O2(g) --> CO2(g) + 2 SO2(g) Reactants: CS2(l) + 3 O2(g) Products: CO2(g) + 2 SO2(g)

CS2(l) + 3 O2(g) --> CO2(g) + 2 SO2(g) 1)Take the first equation and flip it so that CS2 is on the reactant side. The sign of DH changes. 2) Equation 2 does not need to be changed because it has CO2 on the product side. CS2(l) --> C(s) + 2 S(s) DH = - 88 kJ/mol C(s) + O2(g) --> CO2(g) DH = -394 kJ/mol

CS2(l) + 3 O2(g) --> CO2(g) + 2 SO2(g) 3) Equation 3 has SO2 on the product side, but we need to multiply by 2. Multiply DH by 2 also! 2S(s) + 2O2(g) --> 2SO2(g) DH = 2(-297 kJ) = -594 kJ/mol

Add all three equations together and cancel the compounds that are the same but on opposite sides of the arrow. Add the DH values and you will get your final answer. (DH = -1076 kJ/mol)

Calculating DH using Bond Energies

DH = DH bonds broken - DH bonds formed Each type of bond has a characteristic bond energy Breaking bonds requires energy. It is endothermic. Making new bonds gives out energy. It is exothermic. The energy associated for different bonds can be found on p. 156 & p. 419 of your textbook. To calculate DH use the formula: DH = DH bonds broken - DH bonds formed

H2(g) + Br2(g) --> 2HBr(g) Calculate the DH for the above reaction. You need to use the values of page 156 & 419 of your textbook. DH = DH bonds broken - DH bonds formed DH =(436+193) - (366+366) DH = - 103 kJ/mol Bonds Broken: H-H 436 kJ/mol Br-Br 193 kJ/mol Bonds Formed: H-Br 366 kJ/mol H-Br 366 kJ/mol