Question

34. 3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 11.21°C. Calculate ∆H, in kJ/mol, for the dissolution of the unknown solid. (The specific heat of the solution is 4.184 J/g･°C and the density of the solution is 1.20 g/mL).

Answer 1

Answer:

ΔH = 2.68kJ/mol

Explanation:

The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:

q = m*S*ΔT

Where q is heat of reaction in J,

m is the mass of the solution in g,

S is specific heat of the solution = 4.184J/g°C

ΔT is change in temperature = 11.21°C

The mass of the solution is obtained from the volume and the density as follows:

150.0mL * (1.20g/mL) = 180.0g

Replacing:

q = 180.0g*4.184J/g°C*11.21°C

q = 8442J

q = 8.44kJ when 3.15 moles of the solid react.

The ΔH of the reaction is:

8.44kJ/3.15 mol

= 2.68kJ/mol