Asked by LaurenB
Compute the change in temperature when 1.1 g of sodium chloride is dissolved in 200.0 mL of water initially at 28 C. The heat capacity of water is 4.184 J/mL/K. Neglect the heat capacity of the NaCl.
Species ΔfH (kJ/mol)
NaCl(aq)-407.3
NaCl(s) -411.2
Species ΔfH (kJ/mol)
NaCl(aq)-407.3
NaCl(s) -411.2
Answers
Answered by
DrBob222
DH = delta H.
DHsoln = DHsolvation - DHlatticeenergy
DHsoln = -407.3 -(-411.2) = 3.9 kJ/mol
The + means it is an endothermic reaction and the water will become cooler; we are extracting heat from the water. How much? 3.9 kJ/mol x (1000 J/kJ) x 1.1 g NaCl x (1 mol NaCl/molar mass NaCl) = q and since we are extracting heat it is -q.
Then -q = mass H2O x specific heat H2O x (Tfinal-Tinitial) and solve for Tfinal.
DHsoln = DHsolvation - DHlatticeenergy
DHsoln = -407.3 -(-411.2) = 3.9 kJ/mol
The + means it is an endothermic reaction and the water will become cooler; we are extracting heat from the water. How much? 3.9 kJ/mol x (1000 J/kJ) x 1.1 g NaCl x (1 mol NaCl/molar mass NaCl) = q and since we are extracting heat it is -q.
Then -q = mass H2O x specific heat H2O x (Tfinal-Tinitial) and solve for Tfinal.
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