Question

When a chemist mixed 3.60 g of LiOH and 180. mL of 0.65 M HCl in a constant-pressure calorimeter, the final temperature of the mixture was 25.4°C. Both the HCl and LiOH had the same initial temperature, 20.8°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.

Answer 1

The equation for this neutralization reaction is:

LiOH(s) + HCl(aq) ? LiCl(aq) + H2O(l).

Mole of LiOH = 3.60 g/ 23.95 g/mol

= 0.15 moles

Mole of HCl = molarity * volume in L

= 0.65* 180/1000

= 0.117 Moles

Thus LiOH is a limiting agent

The limiting agent has due to following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

Total mass of solution = 180+ 3.60= 183.60 g

Heat asborbed by solution = specific heat x temp change x mass

                = ( 4.18 ) x ( 20.8-25.4) x 183.6

= - 3530 J

= - 3.53 KJ

enthalphy of reaction in KJ/mol = ( -3.53 /0.15) = - 23.535 KJ/mol  

( -ve sign indcates heat is released in reaction)