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In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is
[IIT JEE 2009]
Correct answer is '9'. Can you explain this answer?
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Verified Answer
In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mo...
Energy released by combustion of 3.5 g gas = 2.5×(298.45-298) kJ
Energy released by 1 mole of gas = 2.5×0.45/3.5/28 = 9 kJ mol-1
Energy released by 1 mole of gas = 2.5×0.45/3.5/28 = 9 kJ mol-1
Most Upvoted Answer
In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mo...
Given:
- Mass of gas = 3.5 g
- Molar mass of gas = 28 g/mol
- Temperature change in calorimeter = 0.45 K
- Heat capacity of calorimeter = 2.5 kJ/K
To Find:
The heat given out in kJ/mol
Solution:
Step 1: Calculate the moles of the gas:
Moles of gas = Mass of gas / Molar mass of gas
Moles of gas = 3.5 g / 28 g/mol
Moles of gas = 0.125 mol
Step 2: Calculate the heat absorbed by the calorimeter:
Heat absorbed by the calorimeter = Heat capacity of calorimeter * Temperature change
Heat absorbed by the calorimeter = 2.5 kJ/K * 0.45 K
Heat absorbed by the calorimeter = 1.125 kJ
Step 3: Calculate the heat given out by the gas:
Since the reaction is exothermic (heat is given out), the magnitude of heat given out by the gas will be equal to the magnitude of heat absorbed by the calorimeter.
Heat given out by the gas = Heat absorbed by the calorimeter
Heat given out by the gas = 1.125 kJ
Step 4: Calculate the heat given out in kJ/mol:
Heat given out in kJ/mol = Heat given out by the gas / Moles of gas
Heat given out in kJ/mol = 1.125 kJ / 0.125 mol
Heat given out in kJ/mol = 9 kJ/mol
Answer:
The numerical value of the heat given out in kJ/mol is 9.
- Mass of gas = 3.5 g
- Molar mass of gas = 28 g/mol
- Temperature change in calorimeter = 0.45 K
- Heat capacity of calorimeter = 2.5 kJ/K
To Find:
The heat given out in kJ/mol
Solution:
Step 1: Calculate the moles of the gas:
Moles of gas = Mass of gas / Molar mass of gas
Moles of gas = 3.5 g / 28 g/mol
Moles of gas = 0.125 mol
Step 2: Calculate the heat absorbed by the calorimeter:
Heat absorbed by the calorimeter = Heat capacity of calorimeter * Temperature change
Heat absorbed by the calorimeter = 2.5 kJ/K * 0.45 K
Heat absorbed by the calorimeter = 1.125 kJ
Step 3: Calculate the heat given out by the gas:
Since the reaction is exothermic (heat is given out), the magnitude of heat given out by the gas will be equal to the magnitude of heat absorbed by the calorimeter.
Heat given out by the gas = Heat absorbed by the calorimeter
Heat given out by the gas = 1.125 kJ
Step 4: Calculate the heat given out in kJ/mol:
Heat given out in kJ/mol = Heat given out by the gas / Moles of gas
Heat given out in kJ/mol = 1.125 kJ / 0.125 mol
Heat given out in kJ/mol = 9 kJ/mol
Answer:
The numerical value of the heat given out in kJ/mol is 9.
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In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer?
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In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer?.
In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer?.
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ample number of questions to practice In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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