Question

Part A

A calorimeter contains 35.0 mL of water at 12.5 ∘C . When 2.10 g of X (a substance with a molar mass of 79.0 g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)→X(aq)

and the temperature of the solution increases to 28.0 ∘C .

Calculate the enthalpy change, ΔH, for this reaction per mole of X.

Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.

Express the change in enthalpy in kilojoules per mole to three significant figures.

Answer 1

Answer – Given, volume of water = 35.0 mL, mass of water = 35.0 g, since density = 1.0 g/mL mass of X = 2.10 g

Ti = 12.5^oC , tf = 28.0^oC

We know,

Heat loss = heat gain

Heat gain by water

q = m*C*∆t

    = 35.0 g * 4.184 J/g^oC*(28-12.5)^oC

   = 2270 J

So, heat loss by X = 2270 J

                               = 2.270 kJ

Moles of X = 2.10 g / 79.0 g.mol^-1

                    = 0.0266 moles

So, ∆H = -q

             = -2.270 kJ / 0.0266 moles

            = - 85.4 kJ/mol