need help! :) NOTES: Part 1:calorimeter- 21.5c when 50ml of water added temperat
ID: 972658 • Letter: N
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need help! :)
NOTES:
Part 1:calorimeter- 21.5c when 50ml of water added
temperature of combined water from flask to calorimeter when heated to 60c- 40.1C
Part 2: calorimeter- 21.5c when 25 ml of hydrochloric acid added
highest temperture is 33.2c when 25 ml of ammonia added
Part 3: calorimeter- 21.5c when 25 ml of water added
highest temperature- 9.1c when 5g of ammonia chloride added
Chemistry
Enthalpy Change for the Decomposition of Ammonium Chloride
Part I – Calorimetry
1. Record the following initial conditions:
(a) Volume of cold water in the calorimeter (mL): 20ml
(b) Mass of water (density = 1g/mL): 20 grams
(c) Initial temperature (C): 21.5C
(d) Volume of hot water (mL): 20 ml
(e) Mass of hot water (density = 1g/mL): 20 grams
(f) Initial temperature of hot water(C): 83.03C
2. Calculate the heat lost by the hot water, q(hot) = (mass × Cp × T)(hot water):
3. Calculate the heat gained by the cold water, q(cold) = (mass × Cp × T)(cold water):
4. Calculate the heat gained by the calorimeter, q(calorimeter) = |q(hot)| - |q(cold)|:
5. Calculate the calorimeter constant by dividing the heat gained by the calorimeter (#4) by the temperature change of the calormineter (same as the temperature change of the cold water):
Part II
1. Record and calculate the following for the heat of the reaction:
(a) Total volume in the calorimeter (mL):
(b) Mass of the solution (assuming a density of 1g/mL) (g):
(c) Temperature change from the reaction, delta T (C):
(d) Calculate the enthalpy change (heat) of the solution (water) using qH2O = mCpT, with Cp=4.184 J/g°C (J):
(e) Calculate the enthalpy change (heat) of the calorimeter using qcal = CcalT (J):
(f) Calculate the enthalpy change for the reaction using qrxn = -qH2O - qcal (J):
2. Convert the total heat of reaction to a molar enthalpy:
(a) Moles of HCl in 25 mL=
(b) Enthalpy of the reaction, per mole of HCl (J/mol): Take the corrected heat of reaction and divide it by the moles of HCl present as follows:
(c) Moles of NH3 in 25 mL=
(d) Enthalpy of the reaction, per mole of NH3 (J/mol):
3. Record and calculate the following for the enthalpy of dissolution:
(a) Total mass in the calorimeter (g): 30grams
(b) Temperature change, T (°C): 10.91C
(c) Calculate the enthalpy change (heat) of the solution using qH2O = mCpT, with Cp=4.184 J/g°C (J):
(d) Calculate the enthalpy change (heat) of the calorimeter using qcal = CcalT:
(e) Calculate the enthalpy change for the reaction using qrxn = -qH2O - qcal (J):
4. Convert this to a molar enthalpy:
(a) Moles of NH4 Cl in 5g (MW = 53.492) = 0.0934 moles
(b) Enthalpy of the reaction, per mole of NH4Cl (J/mol): Qreaction/mol of NH4Cl=
Part III
1. Write out the reaction NH4Cl(s) -> NH3 (g) + HCl(g) as a series of steps which include the reactions observed in Procedures 2 and 3.
Use the known enthalpies for the change of state of NH3 and Hcl, given below.
NH3 (g) -> NH3 (aq) (H = -34,640 J/mol)
HCl (g) -> HCl (aq) (H = -75,140 J/mol)
Be sure to show how the reaction steps must proceed so that H for the desired reaction can be calculated. And be careful to use the positive or negative enthalpy values depending on the direction of the reactions that you add together.
Explanation / Answer
Enthalpy Change for the Decomposition of Ammonium Chloride
Part I – Calorimetry
1. Record the following initial conditions:
(a) Volume of cold water in the calorimeter (mL): 20ml
(b) Mass of water (density = 1g/mL): 20 grams
(c) Initial temperature (C): 21.5C
(d) Volume of hot water (mL): 20 ml
(e) Mass of hot water (density = 1g/mL): 20 grams
(f) Initial temperature of hot water(C): 83.03C
2. heat lost by the hot water, q(hot) = (mass × Cp × T)(hot water): 20g*4.18 J/g C*(83.03-40.1)C=3588.95 J
3. heat gained by the cold water, q(cold) = (mass × Cp × T)(cold water):=density*volume*cp*T=20ml*1g/ml*4.18 J/g C*(21.5-40.1)C =-1554.96J
4. heat gained by the calorimeter, q(calorimeter) = |q(hot)| - |q(cold)|:3588.95-(1554.96)=2033.99J
5. Calculate the calorimeter constant by dividing the heat gained by the calorimeter (#4) by the temperature change of the calormineter (same as the temperature change of the cold water):2033.99J/(40.1-21.5)C=109.35J/C
Part II
1. Record and calculate the following for the heat of the reaction:
(a) Total volume in the calorimeter (mL): 50 ml
(b) Mass of the solution (assuming a density of 1g/mL) (g): 50ml*1g/ml=50g
(c) Temperature change from the reaction, delta T (C):33.2-21.5=11.7C
(d) Calculate the enthalpy change (heat) of the solution (water) using qH2O = mCpT, with Cp=4.184 J/g°C (J):
50g*4.184 J/g C*11.7C=2447.64 J
(e) Calculate the enthalpy change (heat) of the calorimeter using qcal = CcalT (J): 109.35J/C*11.7 C=1279.39 J
(f) Calculate the enthalpy change for the reaction using qrxn = -qH2O - qcal (J):-2447.64-1279.39=-1168.24 J
2. Convert the total heat of reaction to a molar enthalpy:
(a) Moles of HCl in 25 mL=
0.025 liters @ 2 mol/litre = 0.050 mole experiment
(b) Enthalpy of the reaction, per mole of HCl (J/mol): Take the corrected heat of reaction and divide it by the moles of HCl present as follows:
-1168.24 J Joules / 0.050 moles = -23364.8 joules per mole HCl
(c) Moles of NH3 in 25 mL=
0.025 liters @ 2 mol/litre = 0.050 mole experiment
(d) Enthalpy of the reaction, per mole of NH3 (J/mol):
-1168.24 J Joules / 0.050 moles = -23364.8 joules per mole NH3
3. Record and calculate the following for the enthalpy of dissolution:
(a) Total mass in the calorimeter (g): 30grams
(b) Temperature change, T (°C): 10.91C
(c) Calculate the enthalpy change (heat) of the solution using qH2O = mCpT, with Cp=4.184 J/g°C (J):30g*4.184 J/gC*10.91C=1369.42 J
(d) Calculate the enthalpy change (heat) of the calorimeter using qcal = CcalT: 109.35J/C*10.91C=1193.0J
(e) Calculate the enthalpy change for the reaction using qrxn = -qH2O - qcal (J): -1369.42 J-1193.0J=-2562.42 J
4. Convert this to a molar enthalpy:
(a) Moles of NH4 Cl in 5g (MW = 53.492) =5g/53.492g/mol= 0.0934 moles
(b) Enthalpy of the reaction, per mole of NH4Cl (J/mol): Qreaction/mol of NH4Cl==-2562.42 J/0.0934 mol=-27434.90 J/mol
Part III
1. Write out the reaction NH4Cl(s) -> NH3 (g) + HCl(g) as a series of steps which include the reactions observed in Procedures 2 and 3.
Use the known enthalpies for the change of state of NH3 and Hcl, given below.
NH3 (g) -> NH3 (aq) (H = -34,640 J/mol)
HCl (g) -> HCl (aq) (H = -75,140 J/mol)
enthalpy of dissolution: NH4Cl(s)àNH4+(aq) +Cl-(aq) H =-27434.90 J/mol……….(1)
HCl(aq)+NH3(aq)-à NH4+(aq) +Cl-(aq) , H=-23364.8 joule/ mole……….(2)
……………………………………………………………………………..
NH3 (g)+ HCl (g) -> NH3 (aq)+ HCl (aq) H=-109,780 J/mol……….(3)
Eqn 1-2-3, gives NH4Cl(s) -> NH3 (g) + HCl(g) [H=-27434.90 –(-23364.8)-( -109,780)=105709.9 J/mol
Be sure to show how the reaction steps must proceed so that H for the desired reaction can be calculated. And be careful to use the positive or negative enthalpy values depending on the direction of the reactions that you add together.
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