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The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd and IIIrd ionization enthalpies are 580, 1820 and 2740 kJ respectively. How much heat has too be supplied (in kJ) to convert 13.5 g of aluminium into Al3+ ions and electrons at 298 k
- a)5470
- b)2735
- c)4105
- d)3765
Correct answer is option 'B'. Can you explain this answer?
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The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd ...
Heat needed too be supplied
per mol = 330 + 580 + 1820 + 2740
= 5470 kJ
Heat required = 0.5 × 5470 kJ = 2735 kJ
per mol = 330 + 580 + 1820 + 2740
= 5470 kJ
Heat required = 0.5 × 5470 kJ = 2735 kJ
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The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd ...
Enthalpy of Sublimation of Aluminium
The enthalpy of sublimation of aluminium is given as 330 kJ/mol. This means that 330 kJ of heat is required to convert one mole of solid aluminium into gaseous aluminium atoms.
Ionization Enthalpies of Aluminium
The ionization enthalpies of aluminium are given as follows:
1st ionization enthalpy = 580 kJ/mol
2nd ionization enthalpy = 1820 kJ/mol
3rd ionization enthalpy = 2740 kJ/mol
Conversion of Aluminium to Al3+ ions and Electrons
To convert aluminium into Al3+ ions and electrons, we need to consider the following steps:
1. Sublimation of solid aluminium to gaseous aluminium atoms: This step requires 330 kJ/mol of heat.
2. Ionization of gaseous aluminium atoms to form Al+ ions: This step requires the first ionization enthalpy of aluminium, which is 580 kJ/mol.
3. Ionization of Al+ ions to form Al2+ ions: This step requires the second ionization enthalpy of aluminium, which is 1820 kJ/mol.
4. Ionization of Al2+ ions to form Al3+ ions: This step requires the third ionization enthalpy of aluminium, which is 2740 kJ/mol.
Calculating the Heat Required
To calculate the heat required to convert 13.5 g of aluminium into Al3+ ions and electrons, we need to use the molar mass of aluminium, which is 26.98 g/mol.
1. Calculate the number of moles of aluminium:
Number of moles = Mass / Molar mass
Number of moles = 13.5 g / 26.98 g/mol
Number of moles = 0.501 moles
2. Calculate the heat required for each step:
Heat required for sublimation = 0.501 moles * 330 kJ/mol = 165.33 kJ
Heat required for 1st ionization = 0.501 moles * 580 kJ/mol = 290.58 kJ
Heat required for 2nd ionization = 0.501 moles * 1820 kJ/mol = 913.22 kJ
Heat required for 3rd ionization = 0.501 moles * 2740 kJ/mol = 1372.74 kJ
3. Calculate the total heat required:
Total heat required = Heat required for sublimation + Heat required for 1st ionization + Heat required for 2nd ionization + Heat required for 3rd ionization
Total heat required = 165.33 kJ + 290.58 kJ + 913.22 kJ + 1372.74 kJ
Total heat required = 2741.87 kJ
Therefore, the correct answer is option B) 2735 kJ.
The enthalpy of sublimation of aluminium is given as 330 kJ/mol. This means that 330 kJ of heat is required to convert one mole of solid aluminium into gaseous aluminium atoms.
Ionization Enthalpies of Aluminium
The ionization enthalpies of aluminium are given as follows:
1st ionization enthalpy = 580 kJ/mol
2nd ionization enthalpy = 1820 kJ/mol
3rd ionization enthalpy = 2740 kJ/mol
Conversion of Aluminium to Al3+ ions and Electrons
To convert aluminium into Al3+ ions and electrons, we need to consider the following steps:
1. Sublimation of solid aluminium to gaseous aluminium atoms: This step requires 330 kJ/mol of heat.
2. Ionization of gaseous aluminium atoms to form Al+ ions: This step requires the first ionization enthalpy of aluminium, which is 580 kJ/mol.
3. Ionization of Al+ ions to form Al2+ ions: This step requires the second ionization enthalpy of aluminium, which is 1820 kJ/mol.
4. Ionization of Al2+ ions to form Al3+ ions: This step requires the third ionization enthalpy of aluminium, which is 2740 kJ/mol.
Calculating the Heat Required
To calculate the heat required to convert 13.5 g of aluminium into Al3+ ions and electrons, we need to use the molar mass of aluminium, which is 26.98 g/mol.
1. Calculate the number of moles of aluminium:
Number of moles = Mass / Molar mass
Number of moles = 13.5 g / 26.98 g/mol
Number of moles = 0.501 moles
2. Calculate the heat required for each step:
Heat required for sublimation = 0.501 moles * 330 kJ/mol = 165.33 kJ
Heat required for 1st ionization = 0.501 moles * 580 kJ/mol = 290.58 kJ
Heat required for 2nd ionization = 0.501 moles * 1820 kJ/mol = 913.22 kJ
Heat required for 3rd ionization = 0.501 moles * 2740 kJ/mol = 1372.74 kJ
3. Calculate the total heat required:
Total heat required = Heat required for sublimation + Heat required for 1st ionization + Heat required for 2nd ionization + Heat required for 3rd ionization
Total heat required = 165.33 kJ + 290.58 kJ + 913.22 kJ + 1372.74 kJ
Total heat required = 2741.87 kJ
Therefore, the correct answer is option B) 2735 kJ.
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The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd and IIIrd ionization enthalpies are 580, 1820 and 2740 kJ respectively. How much heat has too be supplied (in kJ) to convert 13.5 g of aluminium into Al3+ ions and electrons at 298 ka)5470b)2735c)4105d)3765Correct answer is option 'B'. Can you explain this answer?
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The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd and IIIrd ionization enthalpies are 580, 1820 and 2740 kJ respectively. How much heat has too be supplied (in kJ) to convert 13.5 g of aluminium into Al3+ ions and electrons at 298 ka)5470b)2735c)4105d)3765Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd and IIIrd ionization enthalpies are 580, 1820 and 2740 kJ respectively. How much heat has too be supplied (in kJ) to convert 13.5 g of aluminium into Al3+ ions and electrons at 298 ka)5470b)2735c)4105d)3765Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd and IIIrd ionization enthalpies are 580, 1820 and 2740 kJ respectively. How much heat has too be supplied (in kJ) to convert 13.5 g of aluminium into Al3+ ions and electrons at 298 ka)5470b)2735c)4105d)3765Correct answer is option 'B'. Can you explain this answer?.
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