When a chemist mixed 3.60 g of LiOH and 180. mL of 0.65 M HCl in a constant-pres

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.

Explanation / Answer

LiOH moles = mass/Molar mass = 3.6 /23.95 = 0.15

HCl moles = M x V = 0.65 x 180/1000 = 0.117

HCl mass 180 g   ( as density = 1g/ml) , LiOH mass = 3.6 g , total mass = 183.6 g

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

                = ( 4.18 ) x ( 25.4-20.8) 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)

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