Question

Determination of heat capacity of calorimeter	trial 1	trial 2	trial 3
volume of cool water	50.0 mL	50.0mL	50.0mL
volume of warm water	50.0 mL	50.0 mL	50.0 mL
initial temp of cool water=initial calorimeter temp	25.2 C	25.5 C	25.5 C
initial temp of warm water	35.2 C	40.9 C	37.1 C
final temp of water in calorimeter	32.5 C	32.9 C	30.9 C
Enthalpy of Neutralization	Trial 1	trial 2	trial 3
volume of 2.0 M HCl	50.0 mL	50.0 mL	50.0 mL
volume of 2.0 M NaOH	50.0 mL	50.0 mL	50.0 mL
initial temperature HCl	24.5 C	24.9 C	25.1 C
Initial temperature NaOH	25.6 C	25.5 C	25.6 C
Final temperature	35.7 C	35.9 C	35.9 C
Enthalpy of Dissolution of Calcium in Acid	Trial 1	trial 2	trial 3
Mass of Ca(s)	0.5 g	0.5 g	0.5 g
Volume of 1.0 M HCl	50.0 mL	50.0 mL	50.0 mL
initial temperture	25.8 C	27.0 C	26.4 C
final temperature	45.7 C	45.9 C	46.2 C
Enthalpy of dissolution of calcium in water	trial 1	trial 2	trial 3
Mass of Ca(s)	0.5 g	0.5 g	0.5 g
volume of water	50.0 mL	50.0 mL	50.0mL
initial temp	25.6 C	25.7 C	25.6 C
final temp	48.5 C	48.4 C	51.1 C

*Calculate the heat capacity of the calorimeter

*Calculate the enthalpy of the neutralization reaction in kJ/mol

*calculate the enthalphy of dissolution of Ca in acid in kJ/mol

* calculate the enthalphy of dissolution of Ca in water in kJ/mol

Answer 1

1) Calculation of heat capacity of calorimeter:- Trial 1

q = c.T ; c = q / T

Heat given up by warm water = 50(35.2 - 32.5) = 135 cal.

Heat taken up by cold water = 50(32.5 - 25.2) = 365 cal.

Q (calorimeter) = Q (hot water) - Q(cold water)

Heat taken up by calorimeter = 135 - 365 = - 230 cal/⁰C or ⁰K

Heat capacity of calorimeter = -230 / 2.7= - 85.2 cal/°C (or °K) = 85.2 (4.2) = 357.84 J/⁰C

Trial 2 :-

q = c.T ; c = q / T

Heat given up by warm water = 50(40.9 - 32.9) = 400 cal.

Heat taken up by cold water = 50(32.9 - 25.5) = 370 cal.

Heat taken up by calorimeter = 400 - 370 = 30 cal/⁰C or ⁰K

Heat capacity of calorimeter = 30 / 7.4 = 4.05 cal/°C (or °K) = 4.05 (4.2) = 17.01 J/⁰C

Trial 3 :-

Heat given up by warm water = 50(37.1 - 30.9) = 310 cal.

Heat taken up by cold water = 50(30.9 - 25.5) = 270 cal.

Heat taken up by calorimeter = 310 - 270 = 40 cal/⁰C or ⁰K

Heat capacity of calorimeter = 40 / 5.4 = 7.41 cal/°C (or °K) = 31.122 J/⁰C

2) Calculation of enthalpy of neutralisation :- Trial 1(molar enthalpy of neutralisation per mole of HCl)

Chemical equation :- NaOH + HCl → NaCl + H₂O

Moles of HCl = 0.050 L HCl × 2 M HCl / 1L HCl = 0.1 mol HCl

Volume of solution = (50 + 50) mL = 100 mL

Mass of solution = 100 mL ×1.00g / 1mL = 100 g

ΔT = T₂ –T₁ (final - initial) = (35.7 – 24.5) °C = 11.2 °C

Total heat ; Heat from neutralization + Heat to warm solution + Heat to warm calorimeter = 0

q₁ + q₂ + q₃ = 0

n ΔH + m c ΔT+C ΔT = 0

0.1 mol × ΔH + 100 g × 4.184 J·g⁻¹°C⁻¹ × 11.2 °C + 357.84 J°C⁻¹ × 11.2 °C = 0

0.1 mol × ΔH + 4602.4 J + 4.007 J = 0

0.1 mol × ΔH = 4606.41 J

ΔH= −4606.41 J / 0.1mol = - 46064.1 J/mol = -46.064 kJ/mol

Trial 2(molar enthalpy of neutralisation per mole of HCl)

Chemical equation :- NaOH   +   HCl    →     NaCl   +   H₂O

Moles of HCl = 0.050 L HCl × 2 M HCl / 1L HCl = 0.1 mol HCl

Volume of solution = (50 + 50) mL = 100 mL

Mass of solution = 100 mL ×1.00g / 1mL = 100 g

ΔT = T₂ –T₁ (final - initial) = (35.9 – 24.9) °C = 11 °C

Total heat ; Heat from neutralization + Heat to warm solution + Heat to warm calorimeter = 0

q₁ + q₂ + q₃ = 0

n ΔH + m c ΔT+C ΔT = 0

0.1 mol × ΔH + 100 g × 4.184 J·g⁻¹°C⁻¹ × 11 °C + 17.01J°C⁻¹ × 11 °C = 0

0.1 mol × ΔH + 4602.4 J + 187.11 J = 0

0.1 mol × ΔH = - 4789.51 J

ΔH= −4789.51 J / 0.1mol = - 47895.1 J/mol = - 47.895 kJ/mol

Trial 3(molar enthalpy of neutralisation per mole of HCl)

Chemical equation :- NaOH   +   HCl    →     NaCl   +   H₂O

Moles of HCl = 0.050 L HCl × 2 M HCl / 1L HCl = 0.1 mol HCl

Volume of solution = (50 + 50) mL = 100 mL

Mass of solution = 100 mL ×1.00g / 1mL = 100 g

ΔT = T₂ –T₁ (final - initial) = (35.9 – 25.1) °C = 10.8°C

Total heat ; Heat from neutralization + Heat to warm solution + Heat to warm calorimeter = 0

q₁ + q₂ + q₃ = 0

n ΔH + m c ΔT+C ΔT = 0

0.1 mol × ΔH + 100 g × 4.184 J·g⁻¹°C⁻¹ × 10.8 °C + 31.122 J°C⁻¹ × 10.8 °C = 0

0.1 mol × ΔH + 4518.72 J + 336.12 J = 0

0.1 mol × ΔH = - 4854.87

ΔH= −4854.87 J / 0.1mol = - 48548.37J/mol = - 48.55 kJ/mol