Purpose: 

Additiity of reaction heats (Hess’s Law) has been proven through several experiments and calculations.

Materials: 

-graduated cynlinder-plastic foam cup-tongs-balance-thermometer-250ml beaker-goggles-NaOH pallets-tap water-1.0M HCl-1.0M NaOH

Procedure:Part 1: Heat of Solution of Sodium Hydroxide:

1. 200ml of water was measured by graduated cylinder and poured into styrofoam cup.

2. 4.03g of NaOH pallets was obtained from instructor3. Temperature of water in calorimeter was measured to be 21.5°C4. Pallets added to water. Stirring rod was used to dissolve the pallets 

completely.5. Highest temperature of system was recorded to be 27°C

Part 2: Heat of Reaction between Hydrochloric Acid and Sodium Hydroxide Solution:

1. 100ml of 1.0M hydrochloric acid solution was measured by gratuated cylinder and poured into styrofoam cup.

2. 100ml of 1.0M sodium hydroxide solution was measured by gratuated cynlinder and poured into beaker.

3. Temperature was measured for both solutions. Both solutions’s temperature was determined to be 21.0°C.

4. Sodium hydroxide solution was poured into hydrochloric acid solution. Highest temperature recorded to be 21.5°C.

Part 3: Heat of Reaction between Hydrochloric Acid and Solid Soudium Hydroxide:

1. 200ml of 0.5M hydrochloric acid was measured by gratuated cylinder and poured into styrofoam cup.

2. NaOH pallets obtained, mass measured to be 3.97g3. Temperature of hydrochloric acid measured to be 20.0°C4. NaOH pallets added to hydrochloric acid. Stirring rod use to completely 

dissolve the pallets.5. Highest temperature reached by system recorded to be 27.0°C.

Data:

Part 1:

Original Temperature of Water 21.5°CFinal Temperature of Water 27.0°C

Temperature Change 5.5°CMass of 200ml of Water 200gHeat Evolved by Reaction 4598J

Mass of NaOH 4.03gMoles of NaOH 0.1M

Energy per mole of NaOH 45980JΔH1(kJ/mol) NaOH -46.0kJ/mol

Part 2: 

Original temperature of HCl(aq) 21.0°COriginal temperature of NaOH(aq) 21.0°CAverage original temperature 21.0°CFinal temperature of solution 21.5°C

Temperature change 0.5°CTotal mass of solution 200g

Heat evolved by reaction 418JMolarity of NaOH solution  1.0M/LVolume of NaOH solution 0.1L

Moles of NaOH 0.1MEnergy per mole of NaOH 4200J

ΔH2(kJ/mol) NaOH -4.2kJ/mol

Part 3:

Original temperature of HCl(aq) 20.0Final Temperature of solution 27.0

Temperature Change  7.0°CMass of HCl 200g

Heat evolved by reaction 5085JMass of NaOH(s) 3.97gMoles of NaOH 0.1

Energy per mole of NaOH 50850JΔH3(kJ/mol) NaOH -50.85kJ/M

Calculations:

Part 1:

T1 = 21.5°CT2 = 27.0°CΔT = 27.0 – 21.5 = 5.5°Cm = 200gmass of NaOh = 4.03 

ΔH = mΔTc = (200g) (5.5°C) (4.18J/(g*°C)       = 4598Jn = m/M   = 4.03g/40g   = 0.1mol

Energy per mole of NaOH = ΔH/n = 4598J/ 0.1mol = 45980J = 46kJ

ΔH1(kJ/mol) NaOH = -46kJ/mol

Part 2 

T1(HCl)= 21.0°CT1(NaOH) = 21.0°CAverage T1 = 21.0°CT2 = 21.5°CΔT = 21.5 – 21.0 = 0.5°Cm = 200gMolarity of NaOH = 1mol/LVolume of NaOH = 100ml = 0.1L

ΔH = mΔTc = (200g) (0.5°C) (4.18J/(g*°C)       = 420J

n = cv   = (1mol/L)(0.1L)   = 0.1mol

Energy per mole of NaOH = ΔH/n = (420J)/(0.1mol)

= 4200JΔH2(kJ/mol) NaOH = -4.2 kJ

Part 3

T1 = 20°CT2 = 27ΔT = 27 - 20 = 7°Cm = 200gmass of NaOH = 3.97g 

ΔH = mΔTc = (200g) (7.0°C) (4.18J/(g*°C)       = 5085Jn = m/M   = 4.03g/40g   = 0.1mol

Energy per mole of NaOH = ΔH/n = 5085/ 0.1mol = 50850J

ΔH3(kJ/mol) NaOH = -50.85kJ/mol

Questions:

1. When the ethalpy change for all three reactions are compared, one can see all reactions are exothermic. They all release a certain amount of energy throughout the reaction.

2. (2) OH-(aq) + Cl- (aq) -> H2O(l)

(3) NaOH(s) + H- (aq) -> H2O(l) + Na+

(aq)

3. The value of ΔH3 is close to the sum of ΔH2 and ΔH1. This is due to the fact that the chemical reaction of part 3 is the sum of the first two reactions.

4. Percent Deviation = [ (ΔH2 + ΔH1- ΔH3 ) / (ΔH3) ] *100 = [ (-4.2kJ/mol+(-46.0kJ/mol)-( -50.85kJ/M) ) / -50.85kJ/M ] * 100                           = 1.28 %

5. Several reasons for the percent deviation are the inacuracy of the measuring equiptment, assumptions made during calculations such using the specific heat capacity of water for all solutions, the poor insulation of styrofoam cups, and mechanical energy conversion to heat energy when stirring solutions.

6. The energy released in the first reaction would be doubled if the mass of NaOH is doubled hypothetically. This would not effect ΔH1, as the amount of energy is calculated by moles.

Follow Up Question:

1. The reaction between ammonium nitrate and water is extremely endothermic and useful in coldpacks. The coldpack would be install with a containter of ammoium nitrate. Wen the the container is broken, endothermic reaction causes a draws energy from the surrounding enviroment. The loss of heat can then be used to treat athletic injuries to reduce swelling and pain.

Conclusion:

It has been concluded that the addition of a series of reactions can be used to describe the energy change in an overall reaction. ΔHr = ΔH1 + ΔH2 + …