Scheme for Heat of Neutralization & Combustion

Scheme for Patterning Exercise (Heat of neutralization and Heat of combustion) A. Determine the Heat of neutralization (Heat capacity of solution: 4.2 J g-1 0C-1) Volume (cm3) and concentratio n (mol dm-3) 50 cm3 of 1.0 mol dm-3 HCL + 50 cm3 of 1.0 mol dm-3 NH3

Temperature (Ө) @ T:

Number of moles H2O produces: n= MV

Heat given out: H= mc Ө (J)

Initial temperature 29.00C final temperature 35.50C

Number of moles ions H+ (HCl) = 1.0 mol dm-3 X 50 dm-3 1000 = 0.05 mol #

Total volume of mixture = (50 + 50) cm3 = 100 cm3

Temperature change (Ө): (35.5-29.0) 0C Ө = 6.50C

Number of moles ions OH(NH3) = 1.0 mol dm-3 X 50 dm-3 1000 = 0.05 mol #  Therefore, 0.05 mole of water produce

Mass of solution = 100 cm3x 1g cm-3 = 100 g H= mc Ө (J) = 100g x 4.2 J g-1 0 -1 C x 6.50C = 2730 J #

Heat of neutralization: ∆H= H/1000n (kJ mol-1) 0.2 mole of water produces 10080 J. Therefore, 1 mole of water produces: ∆H= H/1000n (kJ) ∆H= 2730 J x 1 mol 0.05 mol (1000) ∆H= 5.46 kJ mol-1 #

Draw energy level diagram: Energy H+(aq)+OH-(aq) ∆H = -5.46 kJ mol-1 H2O(l)

Volume (cm3) and concentratio n (mol dm-3) 150 cm3 of 2.0 moldm-3 HNO3 + 250 cm3 of 1.0 moldm-3 KOH

Temperature (Ө) @ T:

Number of moles H2O produces: n= MV

Heat given out: H= mc Ө (J)

Temperature change (Ө)? Ө = 6.50C

Number of moles ions H+ (HNO3) = 2.0 mol dm-3X 150 dm-3 1000 = 0.3 mol #

Total volume of mixture = (150 + 250) cm3 = 400 cm3

Number of moles ions OH(NaOH) = 1.0 mol dm-3X 250 dm-3 1000 = 0.25 mol #  Therefore, 0.25 mole of water produce

Mass of solution = 400 cm3x 1g cm-3 = 400 g H= mc Ө (J) 14175 J = 400g x 4.2 J g-1 0C-1 x Ө 0C Ө = 8.40C

Heat of neutralization: ∆H= H/1000n (kJ mol-1) ∆H = - 56.7 kJmol-1

Draw energy level diagram: Energy

H = ∆Hx1000n H+(aq)+OH-(aq) H = 56.7x0.25x 1000 H = 14175 J ∆H = -56.7 kJ mol-1 H2O(l)

B. Determine the Heat of combustion (Heat capacity of solution: 4.2 J g-1 0C-1) Volume (cm3) and mass (g) 500 cm3 of H2O Mass C2H5OH Before combustion: 120.0g - After combustion: 115.4g

Temperature (Ө) @ T:

Number of moles: n= m/ RMM

Heat given out: Heat of combustion: H= mc Ө (J) ∆H= H/1000n (kJ mol-1)

Draw energy level diagram:

Initial temperature 28.00C final temperature 75.50C

Mass of CH3OH = 120.0g -115.4g = 4.6 g

Mass of water = 500cm3x 1gcm3 = 500g

Energy

Temperature change (Ө): (75.5-28.0) 0C Ө = 47.50C

Molar mass of CH3OH = 12(2)+1(5)+16+1 = 46g mol-1 Number of moles of CH3OH = 4.6 g / 46 g mol-1 = 0.1 mol

H= mc Ө (J) = 500g x 4.2 J g-1 0C-1 x 47.50C = 99750 J #

0.0156 mole of CH3OH produces 99750 J.

C2H5OH(l)+3O2(g) Therefore, 1 mole of CH3OH produces: ∆H= H/1000n (kJ) ∆H= 99750 J x 1 mol 0.1 mol (1000) ∆H= 997.5 kJ mol-1#

∆H= -997.5 kJ mol-1 2CO2(g)+ 3H2O(l)

Volume (cm3) and mass (g) 250 cm3 of H2O Calculate mass of C3H7OH where produces 2016 kJ?

Temperature (Ө) @ T:

Number of moles: n= m/ RMM

Heat given out: Heat of combustion: H= mc Ө (J) ∆H= H/1000n (kJ mol-1)

Draw energy level diagram:

Increase temperature: 25.00C

Molar mass of C3H7OH = 12(3)+1(7)+16+1 = 60g mol-1

Mass of water = 250cm3x 1gcm3 = 250g

Energy

 Mass of C3H7OH? = 60g x 26.25 kJ 2016 kJ = 0.78g #

H= mc Ө (J) = 250g x 4.2 J g-1 0C-1 x 250C = 26250 J #

∆H= 2016 kJ 60g mol-1 given out 2016 kJ of heat of combustion. Mass of C3H7OH needed if 26.25 kJ of heat release?

C3H7OH(l)+9/2 O2(g) ∆H= -26.25 kJ mol-1 3CO2(g)+ 4H2O(l)