When 25.0 mL of 0.25 mol/L LiOH and 25.0 mL of 0.25 mol/L HCl are mixed together, the temperature warms 15.8oC. What is the molar enthalpy of neutralization for LiOH?

Question

Need help solving this, only one temperature present.

Answer 1

q = mC∆T

q = heat = ?

m = mass = 50 g (assuming density of each solution is 1 g/ml as it is for water)

C = specific heat = 5.184 J/gº (assuming a specific heat identical to water)

∆T = change in temperature = 15.8º (assuming it warms "by" 15.8º)

q = (50 g)(4.184 J/gº)(15.8º) = 3305 J

LiOH + HCl ==> LiCl + H2O ... balanced equation

mols LiOH present = 0.025 L x 0.25 mol/L = 0.00625 mols = mols HCl

molar ∆H_{neutralization} = 3305 J / 0.00625 mols x 1 kJ/1000 J = 529 kJ/mol