I just need help with the last 3 questions of the lab, please! Lab Procedure: Ta

Question

I just need help with the last 3 questions of the lab, please!

Lab Procedure:

Take a new calorimeter from the Containers shelf and place it on the workbench.

Take water from the Materials shelf and add 25.0 mL to the calorimeter.

Take a thermometer from the Instruments shelf and attach it to the calorimeter. Record the temperature inside the calorimeter in your Lab Notes.

Take ammonium chloride (NH4Cl) from the Materials shelf and add 5.000 g to the calorimeter.

Observe the temperature change in the calorimeter as the reaction proceeds. Record the new temperature your Lab Notes.

Lab Notes:

Experiment 3: Determine the Enthalpy of Dissolution of NH4Cl in Water

Temp of calorimeter + 25 mL water: 21.5 C

Temp of calorimeter + 25 mL water + 5.000 g ammonium chloride: 9.1 C

5. Calculate the enthalpy of the dissolution of NH4Cl in water. The molar mass of NH4Cl is 53.49 g/mol. To receive full points for this question, list every step of your calculation.

6. Given the data in the table below, what is the enthalpy of reaction for the following sulphur dioxide synthesis reaction?

7. Given the data in the table below, what is the enthalpy of dissolution of KOH? The molar mass of KOH is 56.11 g/mol, the specific heat of solution is 4.184 J/gºC, and the calorimeter constant is 22.45 J/ºC. To receive full points for this question, list every step of your calculation.

Thank you!!

Explanation / Answer

5. Determine the Enthalpy of Dissolution of NH4Cl in Water

Given-

Temp of calorimeter + 25 mL water: 21.50 C

Temp of calorimeter + 25 mL water + 5.000 g ammonium chloride: 9.10 C

The molar mass of NH4Cl is 53.49 g/mol.

Moles of NH4Cl = 5.000g/ 53.49 g/mol = 0.0935 moles

In this dissolution experiment Ammonium chloride dissolves in water and absorbs heat energy from the surrounding water, which causes the temperature of water to go down from 210C to 9.10 C

  The change in temperature is 210C - 9.10 C = 11.90C

Specific heat of water = 4.18 J / g o C

Mass of water = 25 ml x 1g/ml(density of water) = 25 g

Use the formula Q = mass of water x specific heat of water x change in Temperature

Q = 25 g x 4.18 J / g o C x – 11.90C

= - 1243.55 J = -1.244KJ

Water has lost -1.244 kJ of heat energy to the salt, or salt has gained + 1.244kJ of heat energy from water. So energy gained by salt is 1.244 kJ.

Change in enthalpy = Q gained by salt / moles of salt

H = 1.244 kJ / 0.0935 mol

      = + 13.30KJ/mol

Enthalpy of Dissolution of NH4Cl in Water= + 13.3 KJ/mol

7. The enthalpy of dissolution of KOH-

Given-

Mass of water = 50.000 g

Initial temperature of the water              22.0 ºC

Mass of KOH added to the calorimeter 1.824 g

Final temperature in the calorimeter      29.9 ºC

the specific heat of solution is 4.184 J/gºC

Convert the mass of KOH(s) to an amount in moles-

Molar mass of potassium hydroxide = 56.11 g/mol

Mass of KOH added to the calorimeter =1.824 g

Moles of KOH = 1.824 g/56.11g/mol = 0.0325 moles

The change in temperature is 29.90C – 22.00 C = 7.90C

Use the formula Q = mass of water x specific heat of water x change in Temperature

Q = 50.00 g x 4.18 J / g o C x 7.90C

= 1 651.1J = 1.651KJ

   Water has gain 1.651KJ kJ of heat energy to the salt, or salt has lost – 1.651kJ of heat energy from water.

Change in enthalpy = Q lost by salt / moles of salt

H = - 1.651 kJ / 0.0325 mol

      = - 50.8KJ/mol

The enthalpy of dissolution of KOH is - 50.8KJ/mol  

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