Question

A generic solid, X, has a molar mass of 66.0 g/mol. In a constant-pressure calorimeter, 29.5 g of X is dissolved in 299 g of water at 23.00 °C.

X(s) -----> X(aq)

The temperature of the resulting solution rises to 25.70 °C. Assume the solution has the same specific heat as water, 4.184 J/(g·°C), and that there\'s negligible heat loss to the surroundings.

How much heat was absorbed by the solution?

=_____kJ

What is the enthalpy of the reaction?

deltaH_rxn= _____kJ

Answer 1

Answer – Given, molar mass of X = 66.0 g/mol , mass of X = 29.5 g ,

Mass of water = 299 g , initial temp, ti = 23.0^oC

Tf = 25.70^oC , specific heat of X = 4.184 J/g^oC

Mass of solution = 299 g +29.5 g = 328.5 g

We know the formula

Heat q = m*C*∆t

            = 328.5 g * 4.184 J/g^oC * (25.70^oC - 23.0^oC)

           = 3711 J

           = 3.711 kJ

Heat was absorbed by the solution is 3.711 kJ

We know, ∆H = q

So, ∆H = 3.711 kJ

(If answer in kJ/mol then use this one-

Moles of X = 29.5 g / 66.0 g.mol^-1

                   = 0.447 moles

So, ∆H in kJ/mol =3.711 kJ / 0.447 moles

                             = 8.303 kJ/mol )