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The heat of combustion of ethanol determined in a bomb calorimeter is -670.48kcal/mol at 298K. What is ∆H at 298K for the reaction A) -335.24kcal/mol B) -671.08lcal/mol C) -670.48kcal/mol D) 670.48kcal/mol?
Most Upvoted Answer
The heat of combustion of ethanol determined in a bomb calorimeter is ...
Dhyan de wali baat ye hai . ki yha temp hai.. 25 degree yani 100 se kam..mtlb Water ka state liquid hi hoga..gas nhi
C2H6O(l)+3O2(g)=>2CO2(g)+3H20(l)
toh Delta n(g)= -1
delta H=deltaE+deltan(gas)RT
deltaH =-670.48 - (1×2×298)/1000
delta H=-671.081
option B
C2H6O(l)+3O2(g)=>2CO2(g)+3H20(l)
toh Delta n(g)= -1
delta H=deltaE+deltan(gas)RT
deltaH =-670.48 - (1×2×298)/1000
delta H=-671.081
option B
Community Answer
The heat of combustion of ethanol determined in a bomb calorimeter is ...
To calculate ∆H at 298K for the given reaction, we need to use the heat of combustion of ethanol determined in a bomb calorimeter. The heat of combustion (∆Hc) is the amount of heat released when one mole of a substance is burned completely in excess oxygen.
The given heat of combustion of ethanol is -670.48 kcal/mol at 298K. The negative sign indicates that the reaction is exothermic, meaning heat is released during the combustion of ethanol.
To find the ∆H for the reaction at 298K, we need to determine the stoichiometry of the reaction. Assuming the reaction is the combustion of ethanol (C2H5OH), the balanced equation is:
C2H5OH + 3O2 -> 2CO2 + 3H2O
From the balanced equation, we can see that one mole of ethanol produces two moles of carbon dioxide (CO2) and three moles of water (H2O) during combustion.
Since the heat of combustion is given per mole of ethanol, we can determine the heat released for the balanced equation as follows:
∆H = (2 mol CO2 x ∆Hc) + (3 mol H2O x ∆Hc)
= (2 mol CO2 x -670.48 kcal/mol) + (3 mol H2O x -670.48 kcal/mol)
= -1340.96 kcal/mol + (-2011.44 kcal/mol)
= -3352.4 kcal/mol
Therefore, the ∆H at 298K for the given reaction is -3352.4 kcal/mol.
The given heat of combustion of ethanol is -670.48 kcal/mol at 298K. The negative sign indicates that the reaction is exothermic, meaning heat is released during the combustion of ethanol.
To find the ∆H for the reaction at 298K, we need to determine the stoichiometry of the reaction. Assuming the reaction is the combustion of ethanol (C2H5OH), the balanced equation is:
C2H5OH + 3O2 -> 2CO2 + 3H2O
From the balanced equation, we can see that one mole of ethanol produces two moles of carbon dioxide (CO2) and three moles of water (H2O) during combustion.
Since the heat of combustion is given per mole of ethanol, we can determine the heat released for the balanced equation as follows:
∆H = (2 mol CO2 x ∆Hc) + (3 mol H2O x ∆Hc)
= (2 mol CO2 x -670.48 kcal/mol) + (3 mol H2O x -670.48 kcal/mol)
= -1340.96 kcal/mol + (-2011.44 kcal/mol)
= -3352.4 kcal/mol
Therefore, the ∆H at 298K for the given reaction is -3352.4 kcal/mol.
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The heat of combustion of ethanol determined in a bomb calorimeter is -670.48kcal/mol at 298K. What is ∆H at 298K for the reaction A) -335.24kcal/mol B) -671.08lcal/mol C) -670.48kcal/mol D) 670.48kcal/mol?
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The heat of combustion of ethanol determined in a bomb calorimeter is -670.48kcal/mol at 298K. What is ∆H at 298K for the reaction A) -335.24kcal/mol B) -671.08lcal/mol C) -670.48kcal/mol D) 670.48kcal/mol? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The heat of combustion of ethanol determined in a bomb calorimeter is -670.48kcal/mol at 298K. What is ∆H at 298K for the reaction A) -335.24kcal/mol B) -671.08lcal/mol C) -670.48kcal/mol D) 670.48kcal/mol? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The heat of combustion of ethanol determined in a bomb calorimeter is -670.48kcal/mol at 298K. What is ∆H at 298K for the reaction A) -335.24kcal/mol B) -671.08lcal/mol C) -670.48kcal/mol D) 670.48kcal/mol?.
The heat of combustion of ethanol determined in a bomb calorimeter is -670.48kcal/mol at 298K. What is ∆H at 298K for the reaction A) -335.24kcal/mol B) -671.08lcal/mol C) -670.48kcal/mol D) 670.48kcal/mol? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The heat of combustion of ethanol determined in a bomb calorimeter is -670.48kcal/mol at 298K. What is ∆H at 298K for the reaction A) -335.24kcal/mol B) -671.08lcal/mol C) -670.48kcal/mol D) 670.48kcal/mol? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The heat of combustion of ethanol determined in a bomb calorimeter is -670.48kcal/mol at 298K. What is ∆H at 298K for the reaction A) -335.24kcal/mol B) -671.08lcal/mol C) -670.48kcal/mol D) 670.48kcal/mol?.
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