Question

When pure sulfuric acid dissolves in water, much heat is given off. To measure it, 4.90 g sulfuric acid is added to 175 g water, both at 10.0 ºC in coffee-cup calorimeter. The temperature of the mixture rose to 14.9 ºC. Assuming no heat loss and the density and the specific heat of the solution to be the same as that of water, calculate the heat evolved per mole of sulfuric acid.

Answer 1

the heat of dissolution is essentially heat due to the acid dissolving in wate, typically, it will realease energy

so

-Qloss = Qgain

therefore

-Qdissolution = Qwater

Qwater = m*C*(Tf-Ti)

mass of water = 175 g

C = 4.184 J7gC

dT =14.9°C

Qwater = 175*4.184*(14.9) =10909.78 J

so

Qdissolution = -10909.78 J

mol of h2so4 used = mass/MW = 4.90/98 = 0.05 mol of H2SO4

so

Hsolution = Q/n = -10909.78 /0.05 = -218195.6 J/mol = -218.196 kJ/mol

negative, since it is exothermic process