Question

Na-OH H-o-s-o-H H-0. 0 0 HNOo 1. Calorimetry. In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M H2SO4 are mixed. The initial temperature of eac solution was 25.6 0 C. After the reaction, the temperature of the solution was 31.3 o C Calculate how much heat does the reaction generate? kJ kJ 2. Using heat of formation Enthalpy's calculate how much heat does the reaction generate? 3. Using bond energies calculate how much heat does the reaction will generate? kJ Standard Enthalpy of Formation for Atomic and Molecular lons Cats Almo Cuties AH, ma Ani A mal Anies Humo C- +1059 K( ) 2512 Ara All HR HPO Ca 0, -120,9 -1674 33 -1314 - 11025 -1298.7 -559 Tillt filii PIERA SER I d III lii 3 NX BZE TE -518.4 -2197 -2006 cos CO/ M NO ON PO C NL) 612 -2299 -12941 -477 FROG CN Haq -100 HOD 235.8 All standard enthalpy values we 25C 1 mol concentration and more of pressure Dashboard Calendar To Do Notes

Answer 1

#1.

Heat generated during reaction : ( we solution heat capacity ~ water C ; density : 1 g/ ml)
Q = m * C(H2O )* dT = 200 g * 4.184 J/g-C * 5.7 C = 4769.76 J or 4.77 kJ

NaOH (aq) + H2SO4 (aq) <===> NaSO4H (aq) + H2O (l)

#2. form heat of formation

100 mL of 1.0 M solution:

∆rHº= Products ΣHºm − Reactants ΣHºm

∆rHº= {Hº( H2O(l)) +Hº(H+(aq)) +Hº(Na+(aq))+ Hº(SO42-(aq))} − {2*Hº(H+(aq)) +Hº(Na+(aq))+Hº(OH-(aq)) +Hº(SO42-(aq))}

∆rHº= {Hº( H2O(l)) } − {Hº(H+(aq)) +Hº(OH-(aq)) } = ( -285.8 kJ/mol ) - (-229.9kJ/mol ) = - 55.9 kJ/mol

Since 0.1 mol reacted. thus Heat evolved : 5.59 kJ

#3. from bond energy :

∆rHº= new bond formed − Bonds broken

∆rHº= {Hº( 2*O-H) +Hº(O-Na) } − {Hº(O-H) +Hº(O-Na)}

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