Calorimetry Problem:

Question

A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 143 g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 °C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

 

*answer kJ, 3 sig figs

 

Note: I got -1.45 kJ/mol, -1.51 kJ/mol, and 1.51 kJ/mol through different attempts in solving the problem but all are incorrect.

Answer 1

H = enthalpy = mcΔT, where m = mass of water, c=specific heat of water (4.184 J/gC) and ΔT=T_f-T_i

 

Calculate the heat gained by the water

 

H_water=143 g (4/184 J/gC)(24.70-21.00) = 2214 J

 

From the law of conservation of energy, the heat gained by one object must be lost by another.

H_rxn= -H_water = -2214 J (1 kJ/1000 J) = -2.214 kJ

 

H_rxn per mole = -2.214 kJ/2.00 mol = 1.107 kJ/mol

 

Rounded to 3 sig figs: 1.10 kJ/mol