Question

please help ? part A and B

9. (Basically the same reaction as #8 above, without some of the assumptions) In one experiment, a student placed 50.0mL of 1.00M HCI at 25.5°C in a coffee-cup calorimeter. To this was added 50.0mL of 1.00M NaOH solution also at 25.5°C. The mixture was stirred, and the temperature quickly increased to a maximum of 32.2°C. 2pts a Provide a thermochemical reaction, including AH in kJ. (include sign) b. What is the energy (AH) evolved in kJ per mole of HCI? (include sign) Because the solutions are relatively dilute, we can assume their specific heats are close to that of H20, 4.184Jg '°C. However, do not assume the density of the solutions is that of water. The density of 1.00M HCI is 1.20g/mL and 1.00M NaOH is 1.40g/mL. work space: Provide a thermochemical equation.

Answer 1

Q9. (a.) The thermochemical reaction is :

HCl (aq) + NaOH (aq) $\rightarrow$ NaCl (aq) + H₂O (l) : $\Delta$ H = -57.9 kJ

(b.) Volume of 1.00 M HCl = 50.0 mL

mass of 1.00 M HCl = (volume of 1.00 M HCl) * (density of 1.00 M HCl)

mass of 1.00 M HCl = (50.0 mL) * (1.20 g/mL)

mass of 1.00 M HCl = 60.0 g

Volume of 1.00 M NaOH = 50.0 mL

mass of 1.00 M NaOH = (volume of 1.00 M NaOH) * (density of 1.00 M NaOH)

mass of 1.00 M NaOH = (50.0 mL) * (1.40 g/mL)

mass of 1.00 M NaOH = 70.0 g

Mass of solution = (mass of 1.00 M HCl) + (mass of 1.00 M NaOH)

Mass of solution = (60.0 g) + (70.0 g)

Mass of solution = 130.0 g

Heat gained by solution = (mass of solution) * (specific heat of solution) * (final temperature - initial temperature)

Heat gained by solution = (130.0 g) * (4.184 J/g.^oC) * (32.2 ^oC - 25.5 ^oC)

Heat gained by solution = (130.0 g) * (4.184 J/g.^oC) * (6.7 ^oC)

Heat gained by solution = 3644.3 J

Heat lost by reaction = -(Heat gained by solution)

Heat lost by reaction = -(3644.3 J)

Heat lost by reaction = -3644.3 J

Heat lost by reaction = -3.6443 kJ

moles HCl = (concentration HCl) * (volume HCl solution in Liter)

moles HCl = (1.00 M) * (50.0 x 10^-3 L)

moles HCl = 0.0500 mol

$\Delta$H = (Heat lost by reaction) / (moles HCl)

$\Delta$H = (-3.6443 kJ) / (0.0500 mol HCl)

$\Delta$H = -72.9 kJ/mole HCl