A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 187 g of solution. The reaction caused the temperature of the solution to rise from 21.0 to 24.7 ^\circ \rm 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.

A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 187 g of solution. The reaction caused the temperature of the solution to rise from 21.0 to 24.7 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.

Step first :-

Find the given values

Mass of reactant= 2.oo mol

Mass of solution (m ) = 187 g

Intital temperature T₁ = 21.0 °C

Final temperature T₂ = 24.7 °C

Change in temperature ∆T        = T₂- T₁

                                                  =24.7  - 21.0

                                                  =3.7 °C

Specific heat capacity of water Cp =4.184 J/g°C

Heat generated

                                Q = m x Cp x ∆T                 ……….. (1)

Plug the values in this formula we get

                               

                                                           Q = 187 x 4.184 x 3.7 j

                                                           Q=  2894.91 j

Divide by 1000 to convert in Kj we get

                                                                 Q= 2.89 kj

Change in enthalpy

                                                               ∆H = Q / number of moles

                                                               ∆H = 2.89 kj/ 2 mol

                                                               ∆H = 1.45 Kj/ mol

This reaction is realizing the energy so it a exothermic  reaction

And in exothermic reaction ∆H is always negative

So our answer will ∆H =  - 1.45 Kj/ mol