When a chemist mixed 3.33 g of LiOH and 255 mL of 0.62 M HCl in a constant-press

Question

When a chemist mixed 3.33 g of LiOH and 255 mL of 0.62 M HCl in a constant-pressure calorimeter, the final temperature of the mixture was 23.4°C. Both the HCl and LiOH had the same initial temperature, 20.4°C.

The equation for this neutralization reaction is: LiOH(s) + HCl(aq) LiCl(aq) + H2O(l).

Given that the density of each solution is 1.00 g/mL and the specific heat of the final solution is 4.1801 J/g·K, calculate the enthalpy change for this reaction in kJ/mol LiOH. Assume no heat is lost to the surroundings.

Explanation / Answer

LiOH moles = mass of LiOH / Molar mass of LiOH = ( 3.33g) / ( 23.95 g/mol) = 0.139 mol

mass of HCl solution = volume x density = 255 l x 1g/ml = 255g

solution mass = 3.33 + 255 = 258.333 g

temperature change observed = 23.4 - 20.4 = 3 C or 3K

heat released during reaction = heat absorbed by solution

Heat absorbed by solution = specific heat of solution x mass of solution x temp change

= 4.1801 J/gK x 258.33 g x 3K

= 3239.6 J = 3.24 KJ

now heat released per mole LiOH = ( 3.24 KJ) / ( 0.139 mol LiOH) = 23.3 KJ/mol

Thus enthalphy change = - 23.3 KJ/mol LiOH

( -ve sign indicates heat released during reaction)

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