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 154 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. Enter your answer in kilojoules per mole of compound to three significant figures.

Answer 1

Step first :-

Find the given values

Mass of reactant= 2.oo mol

Mass of solution (m ) = 154 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

Plug the values in this formula we get

Q = 154 x 4.184 x 3.7 j
      Q= 2384.04 j

Q= 2.38 kj

Change in enthalpy
                                                               ∆H = Q / number of moles
                           ∆H = 2.38 kj/ 2 mol
                                                               ∆H = 1.19 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 = - 1.19 Kj/ mol