How much heat (kJ) is lost or gained by 253.0 ml of water when 2.84 g of MgSO4 is dissolved in the water?

Question

when 2.84 g of MgSO4 is dissolved in the water ?
For the following reaction to occur



MgSO4 → Mg2+(aq) + SO42−(aq)



the change in enthalpy must be −91.3 kJ/mol

What is the change in temperature ( oC ) for the above process assuming no heat is lost or gained by the surroundings ?

Answer 1

The molecular weight of MgSO4 is 120
so you have =2.84/120 = 0.0237 mole
So the energy will be ΔH = -91.3*0.0237 = -2.16 kJ

Apparently since for the reaction ΔH<0 energy will be released by the dissociation of MgSO4 and thus water will gain energy.

The change in temperature will be
ΔT=Q/(m*c)= -ΔH/(m*c)= -(-2.16)/(253*4.184) = 0.002 C

(Q=-ΔH, since the amount or energy absorbed by water is equal to that released by the reaction)

assuming that the density of water is 1 g/mL.

So the temperature of water will increase by 0.002 C