NEET Exam > NEET Questions > The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO...
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The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried out in a bomb calorimeter. The heat released was 743kJ/mol. The value of ∆H at 300K will be A) -740.5 kJ/mol B) -741.75 kJ/mol C) -743.0 kJ/mol D) -744.25 kJ/mol?
Most Upvoted Answer
The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried...
**Given information:**
- Reaction: NH2CN(s) + 3/2O2(g) → N2(g) + CO2(g) + H2O(l)
- Heat released: 743 kJ/mol
**Calculating ∆H:**
The heat released in the reaction corresponds to the change in enthalpy (∆H) of the reaction. To calculate ∆H, we need to consider the stoichiometry of the reaction and the enthalpy change per mole of reactant.
1. Determine the stoichiometric coefficients:
- 1 mole of NH2CN reacts with 3/2 moles of O2 to produce 1 mole of N2, 1 mole of CO2, and 1 mole of H2O.
2. Calculate ∆H per mole of reactant:
- Since the heat released is 743 kJ/mol, the ∆H per mole of reactant can be calculated by dividing the heat released by the number of moles of NH2CN in the reaction.
- ∆H per mole of reactant = 743 kJ/mol / 1 mole = 743 kJ/mol.
**Answer:**
The value of ∆H at 300K will be -743.0 kJ/mol (option C).
- Reaction: NH2CN(s) + 3/2O2(g) → N2(g) + CO2(g) + H2O(l)
- Heat released: 743 kJ/mol
**Calculating ∆H:**
The heat released in the reaction corresponds to the change in enthalpy (∆H) of the reaction. To calculate ∆H, we need to consider the stoichiometry of the reaction and the enthalpy change per mole of reactant.
1. Determine the stoichiometric coefficients:
- 1 mole of NH2CN reacts with 3/2 moles of O2 to produce 1 mole of N2, 1 mole of CO2, and 1 mole of H2O.
2. Calculate ∆H per mole of reactant:
- Since the heat released is 743 kJ/mol, the ∆H per mole of reactant can be calculated by dividing the heat released by the number of moles of NH2CN in the reaction.
- ∆H per mole of reactant = 743 kJ/mol / 1 mole = 743 kJ/mol.
**Answer:**
The value of ∆H at 300K will be -743.0 kJ/mol (option C).
Community Answer
The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried...
Since it is Bomb Calorimeter is a constant volume Calorimeter (constant volume is isochoric). So the heat measured by such an instrument is equivalent to the change in internal energy
so,
DeltaH=DeltaE+delta n(gas)RT
deltaH=-743+(0.5×8.314×300)/1000
deltaH=-741.75+1.2471
deltaH=-741.75kJ/mol
option B
so,
DeltaH=DeltaE+delta n(gas)RT
deltaH=-743+(0.5×8.314×300)/1000
deltaH=-741.75+1.2471
deltaH=-741.75kJ/mol
option B
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The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried out in a bomb calorimeter. The heat released was 743kJ/mol. The value of ∆H at 300K will be A) -740.5 kJ/mol B) -741.75 kJ/mol C) -743.0 kJ/mol D) -744.25 kJ/mol?
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The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried out in a bomb calorimeter. The heat released was 743kJ/mol. The value of ∆H at 300K will be A) -740.5 kJ/mol B) -741.75 kJ/mol C) -743.0 kJ/mol D) -744.25 kJ/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 reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried out in a bomb calorimeter. The heat released was 743kJ/mol. The value of ∆H at 300K will be A) -740.5 kJ/mol B) -741.75 kJ/mol C) -743.0 kJ/mol D) -744.25 kJ/mol? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried out in a bomb calorimeter. The heat released was 743kJ/mol. The value of ∆H at 300K will be A) -740.5 kJ/mol B) -741.75 kJ/mol C) -743.0 kJ/mol D) -744.25 kJ/mol?.
The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried out in a bomb calorimeter. The heat released was 743kJ/mol. The value of ∆H at 300K will be A) -740.5 kJ/mol B) -741.75 kJ/mol C) -743.0 kJ/mol D) -744.25 kJ/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 reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried out in a bomb calorimeter. The heat released was 743kJ/mol. The value of ∆H at 300K will be A) -740.5 kJ/mol B) -741.75 kJ/mol C) -743.0 kJ/mol D) -744.25 kJ/mol? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The reaction NH2CN(s) + 3/2O2(g) = N2(g) + CO2(g) + H2O(l) was carried out in a bomb calorimeter. The heat released was 743kJ/mol. The value of ∆H at 300K will be A) -740.5 kJ/mol B) -741.75 kJ/mol C) -743.0 kJ/mol D) -744.25 kJ/mol?.
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