Hess\'s Law Lab - Notes PDF

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Hess's Law Lab Dr. Sinclair Friday, September 21, 2018 SCH4U3

Hess’s Law Lab Purpose: Hess’s Law states that the change in enthalpy is the same for reactions with the same initial reactants and products but different processes. The purpose of this lab is to verify this law. Three reactions (where one can be obtained by combining the other two) will be observed, and the change in enthalpy of each reaction will be calculated to prove Hess’s Law.

Hypothesis: The Heat of Reaction A is the difference of that of B and C. No matter the number of steps, the conversion from a set of reactants to a set of products involves the same change in enthalpy. Hess’s Law is correct.

Safety Precautions: Solid sodium hydroxide was used in this lab. Exposure to any water, including water vapour in the air, makes the surface highly basic. Extra caution was put into ensuring that it was not touched with bare hands and was not put directly onto the pan of the electronic balance. All materials and solutions were disposed of in the appropriate acid/base waste container.

Materials: ● ●

Water Solid sodium hydroxide

● ●

Aqueous sodium hydroxide (0.50 mol/L) Hydrochloric acid (0.50 mol/L)

Apparatus: ● ● ● ●

Volumetric flask Scoopula Thermometer Funnel

● ● ●

Graduated cylinder 2 styrofoam cups Wayboat

Coffee Cup Calorimeter Thermometer

Styrofoam cups

Procedure: REACTION A: Solid sodium hydroxide dissolved in water 1. 100 mL of water was placed in a coffee cup calorimeter. 2. The temperature of the water was recorded. 3. Approximately 1.0 g of solid sodium hydroxide was obtained and the actual mass was recorded. 4. The sodium hydroxide was added to the water and the lid was quickly placed on the calorimeter. 5. The mixture was stirred carefully as the sodium hydroxide dissolved and the highest temperature reached was recorded. REACTION B: Sodium hydroxide neutralized with hydrochloric acid 1. 50 mL of 0.5 mol/L HCl (aq) was placed in a coffee cup calorimeter. 2. 50 mL of 0.5 mol/L NaOH (aq) was measured out. 3. The temperature of each of the solutions was measured and the average of the two was used. 4. The NaOH (aq) was poured into the HCl (aq) and the lid was quickly placed on the calorimeter. 5. The solution was stirred and the highest temperature reached was recorded. REACTION C: Solid sodium hydroxide in hydrochloric acid 1. 100 mL of 0.25 mol/L HCl (aq) was created by diluting 50 mL of 0.5 mol/L HCl (aq) with 50 mL of water. 2. Experiment A was repeated using the 100 mL of 0.25 mol/L HCl (aq) instead of the water.

Pre-Lab Questions: 1. A: NaOH (s) + H₂O (l) → NaOH (aq) + H₂O (l) B: NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l) C: NaOH (s) + HCl (aq) → NaCl (aq) + H₂O (l)

2. C - B: NaOH (s) + HCl (aq) + NaCl (aq) + H₂O (l) → NaCl (aq) + H₂O (l) + NaOH (aq) + HCl (aq) NaOH (s) → NaOH (aq) A: NaOH (s) + H₂O (l) → NaOH (aq) + H₂O (l) NaOH (s) → NaOH (aq) Therefore, A = C - B

Observations: Changes In Temperature Observed for Reactions of Different Solutions Involving Liquid Water, Sodium Hydroxide (solid and aqueous) and Hydrochloric Acid Experiment

Reaction A: Solid sodium hydroxide dissolved in water

Reaction B: Sodium hydroxide neutralized with hydrochloric acid

Quantities of Reactants

100. mL of H2O(l) (tap water)

1.08 g of solid sodium hydroxide

50.0 mL of 0.5 mol/L NaOH(aq)

50.0 mL of 0.5 mol/L HCl(aq)

100. mL of 0.25 mol/ L HCl(aq)

0.96 g of solid sodium hydroxide

Description of Reactants

Clear, colourless liquid

Solid, white powdery pellets

Clear, colourless liquid

Clear, colourless liquid

Clear, colourless liquid

Solid, white powdery pellets

Initial temperature of solvent

26.0oC

26.5oC

26.0oC

25.5oC

Highest temperature reached

28.5oC

30.0oC

30.0oC

Description of product

Clear, colourless solution

Clear, colourless solution

Clear, colourless solution

Reaction C: Solid sodium hydroxide in hydrochloric acid

Average: 26.25oC

Analysis: 1. For Reaction A, calculate: a) The number of moles of NaOH used:

n=

m M

,

n=

1.08 g (22.99+16.00+1.01) g/mol

b) The heat gained by the water:

m=dv =(1 g/mL )( 100. mL )=100. g

,

n=0.0270 mol

Q=mc ΔT , Q=(100. g)(4.18 J /g ∙ ° C)(28.5° C−26.0 °C) Q=1045 J ≈ 1.04 kJ c) The Heat of the Reaction per mole of NaOH.

Δ H =−1.045 kJ , Δ H =n(Δ H °) , Δ H ° =

ΔH n

,

ΔH ° =

−1.045 kJ 0.0270mol Δ H °≈−38.7 kJ /mol 2. For Reaction B, calculate: a) The number of moles of NaOH used in the experiment:

n= cv ,

n=(0.5 mol /L)(0.050 L) , n=0.025 mol

b) The heat gained by the solution. (Using the specific heat capacity for water)

Q=mc Δ T , Q=(100 g)( 4.18 J /g ∙ ° C )(30.0 °C−26.25 °C ) Q=1567.5 J ≈ 1.57 kJ c) The Heat of Reaction per mole of NaOH.

Δ H =−1.5675 kJ ,

Δ H =n(Δ H °) , Δ H ° =

ΔH n

,

ΔH° =

−1.5675 kJ 0.025 mol Δ H °≈−63 kJ /mol 3. For Reaction C, calculate: a) The number of moles of NaOH used.

n=

m M

,

n=

0.96 g 40.00 g /mol

,

n=0.024 mol

b) The heat gained by the solution. (Using the specific heat capacity for water)

Q=mc ΔT , Q=(100 g)( 4.18 J /g ∙ ° C )(30.0 °C−25.5 °C ) Q=1881 J ≈1.88 kJ c) The Heat of Reaction per mole of NaOH (s).

Δ H =−1.881 kJ , Δ H =n(Δ H °) , Δ H ° =

ΔH n

,

ΔH° =

−1.881 kJ 0.024 mol Δ H ° ≈−78 kJ /mol

4. Calculate the Heat of Reaction for Reaction A using the results from the other 2 reactions. A (Product) : NaOH (s) + H₂O (l) → NaOH (aq)+ H₂O (l)

B: NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l) C: NaOH (s) + HCl (aq) → NaCl (aq) + H₂O (l)

Δ H °=¿ - 63 kJ/mol Δ H °=¿ - 78 kJ/mol

B (Reversed): NaCl (aq) + H₂O (l) → NaOH (aq) + HCl (aq)

Δ H °=¿

63 kJ/mol

C + B (Reversed) =

NaOH (s) + HCl (aq) → NaCl (aq) + H₂O (l) NaCl (aq) + H₂O (l) → NaOH (aq) + HCl (aq)

NaOH (s) + H₂O (l)→ NaOH (aq) + H₂O (l) (Reversed)

Δ H °=¿

C+B

= - 78 kJ/mol + 63 kJ/mol = - 15 kJ/mol n=

m M

,

n=

1.08 g , n=0.0270 mol 40.00 g /mol

Δ H °=−15 kJ , Δ H =n(Δ H °)

,

Δ H = 0.0270 mol (-15 kJ/mol) = -0.40 kJ

Therefore, the heat of reaction of Reaction A would be -0.40 kJ. 5. Determine the percent difference between the experimental value of ΔH for Reaction A and the value you calculated in the part above. Experimental value of ΔH for Reaction A: Q = mcΔT = (100g)(4.18 J/g*C°)(28.5oC - 26.0oC) = 1.04 kJ Δ H =¿ -Q Δ H = - 1.04 kJ Calculated ΔH for Reaction A: Δ H =¿ -0.40 kJ Percentage difference =

2(−1.04 −(−0.40 )) × 100% = 89% −0.4−1.04

Therefore, the percentage difference of the experimental value of ΔH for Reaction A and the calculated value of ΔH for Reaction A is 89%.

Discussion: 1. Discuss Hess’ Law in terms of the law of conservation of energy and in terms of the three parts of

this experiment. The Law of Conservation of Energy states that energy cannot be created or destroyed. Enthalpy, by definition, is the total internal energy of a system. Since every substance has a specific enthalpy, the same amount of energy is required to transform one substance into another. It does not matter how many steps are taken. Hess’s Law confirms that the initial reactants and final products are the only things determining the change in enthalpy. Hess’s Law is consistent with the Law of Conservation of Energy.

Sources of Error: Errors made while conducting the experiment

How these errors may have affected the final results

Possible corrections for these errors

There was no way to check if the solutions from the previous reaction was fully cleaned out of the calorimeter.

There may have been even a miniscule amount of residue left from the previous reaction, which may have caused the temperature to either increase or decrease, which would have skewed the data.

If possible, a new coffee cup calorimeter should be used for each reaction, so there would be no chance of residue being left behind from previous experiments.

The calorimeter was not airtight. Stirring with the thermometer can make the hole bigger, and the cups were not made to fit perfectly.

Heat could have escaped from the calorimeter. Since the temperature changes were really small, even a small amount would be significant.

Use a professional calorimeter, or install a propeller and seal the edges of the cup.

There may have been slight variations when locating the bottom of the meniscus for volume measurements.

This would result in inaccurate concentrations for some of the acids used, as well as recording values that are higher or lower than the actual value. This would also affect the moles of the reactants and products.

Use a graduated cylinder with more precise measurements, for example, one with a thinner neck, such as a volumetric flask.

Conclusion: According to our results, Hess’s Law cannot be verified because there was a significant difference (89%) between the calculated value and the experimental value of reaction A. However, this may be due to the sources of error listed above. Further experimentation is required to fully prove or disprove Hess’s Law....