Question

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HCl in a coffee cup calorimeter. If both solutions were initially at 35.0°C and the temperature of the resulting solution was recorded as 37.0°C, determine the experimental AHxt (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCI. Assume that no heat is lost to the calorimeter or the surroundings, and that the density of the solution is 1.03 g/mL and the specific heat of the solution is 4.184 J/g.°C.

Answer 1

Just before starting of the reaction in the medium there were, 100 mL 0.300 M HCl and 100 mL 0.300 M NaOH. So, total volume of solution is 200 mL. Density of the solution is 1.03 g.mL^-1. So, the mass of the solution is (200 mL x 1.03 g.mL^-1) = m = 206 g.

Specific heat = S = 4.184 J / (g. ^oC)

Change in temperature = (37.0 - 35.0) ^oC = $\theta$ = 2.0 ^oC

So, heat absorbed by the calorimeter

$Q=ms\theta =(206g \times 4.184J./(g.^oC)\times 2.0^oC)=1723.808J$

base added = 100 mL 0.300 M = (100 x 0.300) mL x M = 30 mL x (mol / L) = 30 mL x (mol / 10³ mL) = 0.03 mol

So, the amount of heat released by the solution was due to this much moles of base. So, for 1 mol of base amount of heat released is = (1723.808 / 0.03) J.mol^-1 = 57460 J.mol^-1 = 57.460 kJ.mol^-1

Since heat is released due to the reaction so,

$\mathbf{\Delta H^0_{rxtn}=-57.460\, kJ.mol^{-1}}$