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 128 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 1

Given values are:

Mass of reactants= 2.00 moles

Mass of solution= 128g

Initial temperature= 21.00 degee C

Final temperature= 24.70 degree C

Change in Temperature(delta T)= 24.70-21.00= 3.70 degree C

Specific heat of Cp= 4.184J/g degree C

Heat generated Q= m*Cp*delta T

= 128*4.184*3.70

= 1981.5424 J

Dividing Q by 1000 we get Q in kJ= 1981.5424/1000= 1.9815 kJ

Change in Enthalphy (delta H)= Q/(no. of moles)

= 1.9815/2= 0.99075 kJ/mol

Since in this reaction heat is released so the reaction is exothermic and delta H is always negative so

delta H= -0.99075