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For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kcal at 300K. Assuming R = 2cal/K/mol, the value of Del(H) for the above reaction is A) 15.8Kcal B) 18.2kcal C) 20kcal D) 16.4kcal?
Most Upvoted Answer
For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kca...
∆H=∆E + ∆nRT
=17000 + 2×2×300
=18200 Cal
=18.2 kCal
=17000 + 2×2×300
=18200 Cal
=18.2 kCal
Community Answer
For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kca...
Understanding the Reaction and Given Data
The reaction given is:
A(g) + 3B(g) ⇌ 3C(g) + 3D(g)
- The change in internal energy (ΔE) is provided as 17 kcal at 300K.
- The universal gas constant (R) is given as 2 cal/K/mol.
Relation Between ΔE and ΔH
For gaseous reactions, the relationship between the change in internal energy (ΔE) and the change in enthalpy (ΔH) can be expressed as:
ΔH = ΔE + ΔnRT
where Δn is the change in the number of moles of gas.
Calculating Δn
- Reactants: 1 mole of A + 3 moles of B = 4 moles
- Products: 3 moles of C + 3 moles of D = 6 moles
Thus,
Δn = moles of products - moles of reactants = 6 - 4 = 2.
Substituting Values
Now plug in the values into the ΔH equation:
ΔH = ΔE + ΔnRT
- ΔE = 17 kcal = 17000 cal (since 1 kcal = 1000 cal)
- R = 2 cal/K/mol
- T = 300 K
- Δn = 2
Calculating the term ΔnRT:
ΔnRT = 2 moles × 2 cal/K/mol × 300 K = 1200 cal.
Final Calculation of ΔH
Now substitute back into the ΔH equation:
ΔH = 17000 cal + 1200 cal = 18200 cal = 18.2 kcal.
Conclusion
Thus, the value of ΔH for the reaction is 18.2 kcal, which corresponds to option B.
The reaction given is:
A(g) + 3B(g) ⇌ 3C(g) + 3D(g)
- The change in internal energy (ΔE) is provided as 17 kcal at 300K.
- The universal gas constant (R) is given as 2 cal/K/mol.
Relation Between ΔE and ΔH
For gaseous reactions, the relationship between the change in internal energy (ΔE) and the change in enthalpy (ΔH) can be expressed as:
ΔH = ΔE + ΔnRT
where Δn is the change in the number of moles of gas.
Calculating Δn
- Reactants: 1 mole of A + 3 moles of B = 4 moles
- Products: 3 moles of C + 3 moles of D = 6 moles
Thus,
Δn = moles of products - moles of reactants = 6 - 4 = 2.
Substituting Values
Now plug in the values into the ΔH equation:
ΔH = ΔE + ΔnRT
- ΔE = 17 kcal = 17000 cal (since 1 kcal = 1000 cal)
- R = 2 cal/K/mol
- T = 300 K
- Δn = 2
Calculating the term ΔnRT:
ΔnRT = 2 moles × 2 cal/K/mol × 300 K = 1200 cal.
Final Calculation of ΔH
Now substitute back into the ΔH equation:
ΔH = 17000 cal + 1200 cal = 18200 cal = 18.2 kcal.
Conclusion
Thus, the value of ΔH for the reaction is 18.2 kcal, which corresponds to option B.
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For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kcal at 300K. Assuming R = 2cal/K/mol, the value of Del(H) for the above reaction is A) 15.8Kcal B) 18.2kcal C) 20kcal D) 16.4kcal?
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For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kcal at 300K. Assuming R = 2cal/K/mol, the value of Del(H) for the above reaction is A) 15.8Kcal B) 18.2kcal C) 20kcal D) 16.4kcal? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kcal at 300K. Assuming R = 2cal/K/mol, the value of Del(H) for the above reaction is A) 15.8Kcal B) 18.2kcal C) 20kcal D) 16.4kcal? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kcal at 300K. Assuming R = 2cal/K/mol, the value of Del(H) for the above reaction is A) 15.8Kcal B) 18.2kcal C) 20kcal D) 16.4kcal?.
For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kcal at 300K. Assuming R = 2cal/K/mol, the value of Del(H) for the above reaction is A) 15.8Kcal B) 18.2kcal C) 20kcal D) 16.4kcal? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kcal at 300K. Assuming R = 2cal/K/mol, the value of Del(H) for the above reaction is A) 15.8Kcal B) 18.2kcal C) 20kcal D) 16.4kcal? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a gaseous reaction, A(g) + 3B(g) = 3C(g) + 3D(g), Del(E) is 17 kcal at 300K. Assuming R = 2cal/K/mol, the value of Del(H) for the above reaction is A) 15.8Kcal B) 18.2kcal C) 20kcal D) 16.4kcal?.
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