Total volume of the solution = (100 + 100) mL = 200 mL. The density of the solut

Question

Total volume of the solution = (100 + 100) mL = 200 mL.

The density of the solution is 1.00 g/mL; therefore, the mass of the solution is (200 mL)*(1.00 g/mL) = 200.0 g.

The solution is an aqueous solution; hence, the specific heat capacity of the solution is given as 4.184 J/g.°C.

Heat involved in the reaction = (mass of solution)*(specific heat capacity of solution)*(change in temperature of the solution) = (200.0 g)*(4.184 J/g.°C)*(25.86 – 22.50)°C = 2811.648 J.

The balanced chemical equation for the reaction is provided. As per the stoichiometric equation,

1 mole NaOH = 1 mole HCl = 1 mole H2O.

Mole(s) of HCl = mole(s) of NaOH = mole(s) of H2O = (100 mL)*(1 L/1000 mL)*(0.500 M) = 0.05 mole.

Heat of neutralization of the reaction = -(heat involved)/(moles of H2O) = -(2811.648 J)/(0.05 mole) = -56232.96 J/mol = -(56232.96 J/mol)*(1 kJ/1000 J) = -56.23292 kJ/mol -56.23 kJ/mol (ans).

Explanation / Answer

10. A quantity of 100 mL of a 0.500 M HCI was mixed with 100 mL of 0.500 M NaOH in a constant pressure calorimeter of negligible heat capacity. The initial temperature of the HCI and NaOH solutions was the same: 22.50 degrees C and the final temperature of the mixed solutions was 25.86 degrees C. Calculate the heat of neutralization reaction on a molar basis (kJ/mol): NaOH(aq) + HCI(aq) NaCl(aq) + H2O(l) Assume the densities and specific heats of the solutions are the same as for water 1.00 g/mL and 4.184 J/g C

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