When a chemist mixed 3.61 g of LiOH and 170. mL of 0.68 M HCl in a constant-pres

Question

When a chemist mixed 3.61 g of LiOH and 170. mL of 0.68 M HCl in a constant-pressure calorimeter, the final temperature of the mixture was 25.3°C. Both the HCl and LiOH had the same initial temperature, 20.5°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. kJ/mol LiOH

Explanation / Answer

Taking values from above and using,

Heat q = mCpdT

            = 170 x 4.1801 x (25.3 - 20.5)

            = 3410.9616 J

enthalpy change = -3.41096 kJ x 23.95 g/mol/3.61 g = -22.63 kJ/mol LiOH

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