I had to do this online lab and have a total of 8 questions at the end and I am stuck half way through. Can someone please walk me through the last few question?
Here is the Lab and my homework:
In this experiment, we will determine the heat of reaction for a neutralization reaction. We will then compare this to the theoretical heat of neutralization. Prior to starting, make sure you know what the products are off a neutralization reaction (general classification = double displacement reaction). Also know how to write a net ionic equation.
Obtain a calorimeter and add 10mLs of 6.00M sulfuric acid. Obtain a burette and add 50mLs of 6.00M sodium hydroxide. Right click on the calorimeter and select pH meter and collect titration data. Then right click again and choose view titration data. Start the titration by adding NaOH and watch the temperature. Continue the titration until 30.0mLs of NaOH has been added.
Using the graph, determine the maximum temperature of the reaction. (57.9721C) Use this temperature to calculate the heat absorbed by the water. Also note how many mLs of NaOH were required to reach this temperature.(19.8mL NaOH)
At the highest temperature of this titration the Volume was19.8 mL, The Ph levels was 1.4252 pH and the temperature was 58.2867C
When I stopped the titration with 30mL of NaOH the Volume was 30 mL, The Ph levels was 14.1532 pH and the temperature was 49.3241C
Calculate the experimental heat of reaction using qwater = - qreaction. (use the mass of the solution at maximum temperature to be equal to the mass of the water). (57.97C=-57.97C)
Calculate the theoretical heat of reaction. Steps..
1. Write a balanced equation, including the phase labels for each substance.
H2SO4(aq) + 2NaOH(aq) --> Na2SO4(aq) + 2HOH(l)
2. Write the net ionic equation for this reaction. This allows us to ignore the spectator ions when calculating the heats of reactions using standard enthalpies of formation.
H+ + HSO4- + 2OH- --> + SO4^2- + 2HOH(l)
3. Look up the standard enthalpies of formation for each reactant and product.
H+=0
HSO4-= -887.3
2OH= -230
SO4^2-= -909.3
2HOH(l)= ???
4. Using Hess’s law, calculate the standard enthalpy for the reaction. This answer should be in kilojoules or Joules.
5. Divide the heat of reaction by the number of moles of OH- shown in the balanced chemical equation. This gives the heat evolved per 1 mole of hydroxide ion.
6. Calculate the actual number of moles of OH- used in your reaction using the formula for molarity (M = n/V). Make sure you units cancel.
7. Multiply the number of moles of OH- by the conversion factor for heat / mole. This gives you the theoretical heat of reaction for part 2.
8. Calculate the % difference between the theoretical heat of reaction and the experimental heat of reaction. Explain any differences greater than 10%.
And those last questions are?
Calculate the experimental heat of reaction using qwater = - qreaction. (use the mass of the solution at maximum temperature to be equal to the mass of the water). (57.97C=-57.97C)
Calculate the theoretical heat of reaction. Steps..
1. Write a balanced equation, including the phase labels for each substance.
H2SO4(aq) + 2NaOH(aq) --> Na2SO4(aq) + 2HOH(l)
2. Write the net ionic equation for this reaction. This allows us to ignore the spectator ions when calculating the heats of reactions using standard enthalpies of formation.
H+ + HSO4- + 2OH- --> + SO4^2- + 2HOH(l)
3. Look up the standard enthalpies of formation for each reactant and product.
H+=0
HSO4-= -887.3
2OH= -230
SO4^2-= -909.3
2HOH(l)= ???
4. Using Hess’s law, calculate the standard enthalpy for the reaction. This answer should be in kilojoules or Joules.
5. Divide the heat of reaction by the number of moles of OH- shown in the balanced chemical equation. This gives the heat evolved per 1 mole of hydroxide ion.
6. Calculate the actual number of moles of OH- used in your reaction using the formula for molarity (M = n/V). Make sure you units cancel.
7. Multiply the number of moles of OH- by the conversion factor for heat / mole. This gives you the theoretical heat of reaction for part 2.
8. Calculate the % difference between the theoretical heat of reaction and the experimental heat of reaction. Explain any differences greater than 10%
q(water) = mass H2O x specific heat x delta T.
The enthalpy of formation for H2O(l) is -285.8 kJ/mol according to the tables I used.
Follow the instructions for the rest of the problem.
To help you with the last few questions, let's break down the steps and explanations for each one:
Step 4: Using Hess's law, calculate the standard enthalpy for the reaction. This answer should be in kilojoules or Joules.
To calculate the standard enthalpy using Hess's law, you need to use the enthalpies of formation for each reactant and product. These values represent the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states.
For example, in the given equation: H2SO4(aq) + 2NaOH(aq) --> Na2SO4(aq) + 2H2O(l)
- The enthalpy of formation for H2SO4(aq) is not provided, so you will need to look it up. Let's assume its value is -1000 kJ/mol.
- The enthalpy of formation for Na2SO4(aq) is not provided, so you will need to look it up. Let's assume its value is -1500 kJ/mol.
- The enthalpy of formation for NaOH(aq) is not provided, so you will need to look it up. Let's assume its value is -800 kJ/mol.
- The enthalpy of formation for H2O(l) is not provided, so you will need to look it up. Let's assume its value is -500 kJ/mol.
Now, write the balanced equation including the enthalpies of formation for each reactant and product:
H2SO4(aq) + 2NaOH(aq) --> Na2SO4(aq) + 2H2O(l)
(-1000 kJ/mol) + 2*(-800 kJ/mol) = (-1500 kJ/mol) + 2*(-500 kJ/mol)
Simplifying the equation:
-1000 kJ/mol - 1600 kJ/mol = -1500 kJ/mol - 1000 kJ/mol
-2600 kJ/mol = -2500 kJ/mol
So, the standard enthalpy change for the reaction is -2600 kJ/mol.
Step 5: Divide the heat of reaction by the number of moles of OH- shown in the balanced chemical equation. This gives the heat evolved per 1 mole of hydroxide ion.
From the balanced chemical equation, you can see that 2 moles of OH- are involved in the reaction. The heat evolved per mole of OH- is then calculated by dividing the heat of reaction obtained in Step 4 by 2.
Step 6: Calculate the actual number of moles of OH- used in your reaction using the formula for molarity (M = n/V). Make sure your units cancel.
You need to calculate the molarity of the NaOH solution at the given volume of 19.8 mL. Convert the volume from mL to L (0.0198 L). Use the formula M = n/V, where M is the molarity, n is the number of moles, and V is the volume.
Assuming the molarity of NaOH is 6.00 M, you can calculate the number of moles using the formula:
6.00 M = n / 0.0198 L
Rearrange the equation to solve for n:
n = 6.00 M * 0.0198 L
Calculate n to find the actual number of moles of OH- used in your reaction.
Step 7: Multiply the number of moles of OH- by the conversion factor for heat/mole. This gives you the theoretical heat of reaction for part 2.
Multiply the number of moles of OH- obtained in Step 6 by the heat evolved per mole of OH- obtained in Step 5. This will give you the theoretical heat of reaction for part 2.
Step 8: Calculate the % difference between the theoretical heat of reaction and the experimental heat of reaction. Explain any differences greater than 10%.
To calculate the percentage difference, use the formula:
% difference = [(theoretical value - experimental value) / theoretical value] * 100
Calculate the percentage difference between the theoretical heat of reaction obtained in Step 7 and the experimental heat of reaction (-57.97°C).
Examine the percentage difference and explain any differences greater than 10%. This means evaluating whether the experimental value is significantly different from the theoretical value.
---
I hope this breakdown helps you understand how to approach these last few questions in your lab report. Remember to double-check the values and calculations using the given data for the respective enthalpies of formation and molarity of NaOH.