Please HELP - Enthelpy Change of Neutralization Reaction

First, we need to consider the molar mass of sodium (Na) in the equation, as we are now using solid sodium instead of solid NaOH. Additionally, we need to account for the fact that solid sodium will react with water to form sodium hydroxide (NaOH) and hydrogen gas (H2). Therefore, the new thermochemical equation would be: 2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g) This equation takes into consideration the molar mass of sodium and accounts for the formation of NaOH and H2. In summary, the change in temperature
  • #1
borderline182
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here's the procedure and calculations to a lab i just did... the questions i need help answering are at the bottom.

Part 1
1. rinse the graduated cylinder w/ a small quantity of 1.00mol/L NaOH(aq) then add 50mL of 1.00mol/L NaOH(aq) to the coffee cup calorimeter. Record the initial temp.
2. 2. Rinse the graduated cylinder with tape water and then with a small quantity of 1.00mol/L HCl(aq). Quickly and carefully add 50mL of 1.00mol/L HCl(aq) to the NaOH(aq) in the calorimeter.
3. Record the highest temp observed, stirring gently and continuously.

Part 2
1. rinse the graduated cylinder with a small quantity of 1.00mol/L HCl(aq) quickly and carefully add 50mL of 1.00mol/L HCl(aq) to the calorimeter and record the initial temp
2. quickly and carefully measure out 2g of NaOH pellets. Add these to the 1.00mol/L HCl(aq) as soon as possible
3. stir the pellets until they have completely dissovled and record the highest temp.

Calculations

Reaction 1
Initial temp=21.5
Final temp=28
Change in temp= 6.5

Qsol’n = mct
Q = (100)(4.184)(6.5)
Q = 2719.6 J
Qrxn = -2719.6 J

Reaction 2
Initial temp= 21.5
Final temp = 43
Change in temp = 21.5

Qsol’n=mct
Q = (50)(4.184)(21.5)
Q = 4496.8 J
Qrxn = -4497.8 J

# of mol’s of HCl used = 0.05
# of mol’s of NaOH used = 0.05


QUESTIONS I NEED HELP ANSWERING 

WHY was the change in temperature much more significant when using solid NaOH? And why is the enthalpy change (Qrxn) much lower?

What changes do you need to make to the thermochemical equation if you perform the investigation using solid sodium?
 
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  • #2



I would like to clarify some concepts and provide explanations for the questions mentioned in the forum post.

Firstly, let's discuss the concept of enthalpy change (ΔH). Enthalpy is a measure of the total energy of a system, including both its internal energy and the energy required to perform work. In a chemical reaction, the change in enthalpy (ΔH) is a measure of the heat released or absorbed during the reaction.

In the given experiment, we are investigating the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH). In the first reaction, we are using a solution of NaOH, while in the second reaction, we are using solid NaOH pellets.

When using a solution of NaOH, the reaction takes place between the NaOH ions and the HCl ions, resulting in a release of heat. This heat is absorbed by the solution, resulting in a change in temperature. On the other hand, when using solid NaOH pellets, the reaction takes place between the solid NaOH and the HCl solution, resulting in a much larger amount of heat being released. This is because the solid NaOH has a higher concentration of ions compared to the NaOH solution, leading to a more significant reaction and a larger change in temperature.

Now, let's address the questions mentioned in the forum post.

1. Why was the change in temperature much more significant when using solid NaOH?

As mentioned earlier, the change in temperature is more significant when using solid NaOH because of the higher concentration of ions in the solid compared to the solution. This leads to a more significant reaction and a larger amount of heat being released.

2. Why is the enthalpy change (Qrxn) much lower?

The enthalpy change (Qrxn) is lower when using solid NaOH because we are using a smaller amount of solid compared to the solution. In the first reaction, we used 50mL of 1.00mol/L NaOH solution, while in the second reaction, we only used 2g of solid NaOH. This results in a smaller amount of heat being released, leading to a lower enthalpy change.

3. What changes do you need to make to the thermochemical equation if you perform the investigation using solid sodium?

If we were to perform the investigation using solid sodium, there
 
  • #3


The change in temperature was more significant when using solid NaOH because it requires more energy to break down the bonds between the solid particles and dissolve them in the solution. This results in a larger change in temperature compared to using already dissolved NaOH. The enthalpy change (Qrxn) is lower because the reaction is not happening between two aqueous solutions, but between a solid and a solution. This means that there is less heat released or absorbed during the reaction.

To perform the investigation using solid sodium, the thermochemical equation would need to be adjusted to include the enthalpy of fusion for solid sodium. This would account for the energy needed to melt the solid sodium before it can react with the solution. The equation would now be:

Na(s) + HCl(aq) → NaCl(aq) + H2O(l)

ΔHrxn = ΔHfus + ΔHsol’n + ΔHsol’n

Where ΔHfus is the enthalpy of fusion for solid sodium.
 

What is a neutralization reaction?

A neutralization reaction is a chemical reaction between an acid and a base that results in the formation of a salt and water. The acid and base neutralize each other's properties, resulting in a neutral solution.

How does a change in enthalpy affect a neutralization reaction?

A change in enthalpy, or heat, affects a neutralization reaction by determining whether the reaction is exothermic or endothermic. Exothermic reactions release heat, while endothermic reactions absorb heat. This can impact the rate and direction of the reaction.

What factors can affect the enthalpy change in a neutralization reaction?

The enthalpy change in a neutralization reaction can be affected by the strength of the acid and base, the concentration of the reactants, and the temperature of the reaction. Other factors such as pressure and catalysts can also play a role.

What are some real-life applications of neutralization reactions?

Neutralization reactions have many practical applications, such as in the production of fertilizers, cleaning products, and pharmaceuticals. They are also used in wastewater treatment to neutralize acidic or basic solutions before they are released into the environment.

Are there any safety precautions to consider when conducting a neutralization reaction?

Yes, it is important to handle acids and bases with caution as they can be corrosive and cause burns. It is also important to wear protective gear and work in a well-ventilated area. Additionally, neutralization reactions involving strong acids and bases can produce a lot of heat and potentially cause explosions, so proper precautions should be taken.

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