Answer to Question #69316 in General Chemistry for Aj

Question

When a chemist mixed 3.54 g of LiOH and 260. mL of 0.65 M HCl in a constant-pressure calorimeter, the final temperature of the mixture was 24.7°C. Both the HCl and LiOH had the same initial temperature, 21.6°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

LiOH(s) + HCl(aq) → LiCl(aq) + H2O(l)
Mr(LiOH) = 24 g/mol
m(LiOH) = 3.54 g = 3.54/24 = 0.1475 mol
T2-T1 = 24.7 - 21.6 = 3.1 oK
m(sol) = 260 g

Q = Cm(T2-T1)
Q = 4.1801 * 260 * 3.1 = 3368.16 J

0.1475 mol (LiOH) - 3368.16 J
1 mol (LiOH) - x J

x = 22841.8 J

dH = - 22841.8 J