When 1.3584g of sodium acetate trihydrate was mixed into 100.0cm 3 of 0.2000 M H

Question

When 1.3584g of sodium acetate trihydrate was mixed into 100.0cm3 of 0.2000 M HCl (aq) at 25 0Cin a solution calorimeter, its temperature fell by 0.3970C on accoun of the reaction: H3O+ (aq) +NaCH3CO2 . 3H2O (s) -->Na+ (aq) + CH3COOH (aq) + 4 H2O(l). The heat capacity of the calorimeter is 91.0 J/K and the heatcapacity density of the acid solution is 4.144J/Kcm. Determine thestandard emthalpy of formation of the aqueous sodium cation. Thestandard enthalpy of formation of sodium acetate trihydrate is-1064kJ/mol. When 1.3584g of sodium acetate trihydrate was mixed into 100.0cm3 of 0.2000 M HCl (aq) at 25 0Cin a solution calorimeter, its temperature fell by 0.3970C on accoun of the reaction: H3O+ (aq) +NaCH3CO2 . 3H2O (s) -->Na+ (aq) + CH3COOH (aq) + 4 H2O(l). The heat capacity of the calorimeter is 91.0 J/K and the heatcapacity density of the acid solution is 4.144J/Kcm. Determine thestandard emthalpy of formation of the aqueous sodium cation. Thestandard enthalpy of formation of sodium acetate trihydrate is-1064kJ/mol.

Explanation / Answer

Moles of sodium acetate trihydrate = mass/ molar mass                                                   = 1.3584 g / 136.08g/mol                                                   = 0.01 moles . moles of H3O+ = molarity of HCl = molarity * volume                      = 0.2 M * 0.1 L                      = 0.02 moles. . From the equation: H3O+ (aq) +NaCH3CO2 . 3H2O (s) -->Na+ (aq) + CH3COOH (aq) + 4 H2O(l). 1 mole of H3O+ reacts with 1 mole of sodium acetate giving 1 moleof Na+. Hence 0.01 moles of H3O+ will react with 0.01 moles of sodiumacetate to give 0.01 moles of Na+. . Heat given out by the reaction (formation of 0.01 moles of Na+)=                                                                        Heat absorbed by calorimeter + heatabsorbed by solution.                                     = (91.0 J.K * 0.397 K)+ (100 cm^3 * 4.144 J/K.cm^3 * 0.397 K)                                     = 200.64 J                            {Assuming heatcapacity of acid solution = 4.144 J/K.cm^3}                                     = 0.2 kJ . Heat of formation of sodium cation = 0.2 kJ / 0.01 moles                                                   = 20 kJ/ mol

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