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Calculate the enthalpy change for the process CCI4 (g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4 (g) Δvap, H° (CCl4) = 30.5 kJ mol -1 ΔfH0 (CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1, where ΔaH° is enthalpy of atomisation Δa H°(Cl2)= 242 kJ mol-1
- a)397 kJ mol−1
- b)345 kJ mol−1
- c)307 kJ mol−1
- d)327 kJ mol−1
Correct answer is option 'D'. Can you explain this answer?
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Calculate the enthalpy change for the process CCI4(g) C (g) → 4 C...
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Calculate the enthalpy change for the process CCI4(g) C (g) → 4 C...
The enthalpy change for the process CCl4(g) -> C(g) can be calculated using the bond enthalpy method.
First, we need to break the bonds in CCl4(g) and form the bonds in C(g).
Breaking the bonds in CCl4(g):
CCl4(g) -> C(g) + 4Cl(g)
The bond enthalpy values for C-Cl bond and C-C bond are as follows:
C-Cl bond enthalpy = 328 kJ/mol
C-C bond enthalpy = 347 kJ/mol
Breaking the C-Cl bonds in CCl4(g) would require 4 * 328 kJ/mol = 1312 kJ/mol.
Forming the bond in C(g):
C(g) -> 1/2 C2(g)
The bond enthalpy value for C-C bond is 347 kJ/mol. Forming the bond in C(g) would release 1/2 * 347 kJ/mol = 173.5 kJ/mol.
Now, we can calculate the enthalpy change for the process:
Enthalpy change = (Energy required to break bonds) - (Energy released by forming bonds)
= 1312 kJ/mol - 173.5 kJ/mol
= 1138.5 kJ/mol
Therefore, the enthalpy change for the process CCl4(g) -> C(g) is 1138.5 kJ/mol.
First, we need to break the bonds in CCl4(g) and form the bonds in C(g).
Breaking the bonds in CCl4(g):
CCl4(g) -> C(g) + 4Cl(g)
The bond enthalpy values for C-Cl bond and C-C bond are as follows:
C-Cl bond enthalpy = 328 kJ/mol
C-C bond enthalpy = 347 kJ/mol
Breaking the C-Cl bonds in CCl4(g) would require 4 * 328 kJ/mol = 1312 kJ/mol.
Forming the bond in C(g):
C(g) -> 1/2 C2(g)
The bond enthalpy value for C-C bond is 347 kJ/mol. Forming the bond in C(g) would release 1/2 * 347 kJ/mol = 173.5 kJ/mol.
Now, we can calculate the enthalpy change for the process:
Enthalpy change = (Energy required to break bonds) - (Energy released by forming bonds)
= 1312 kJ/mol - 173.5 kJ/mol
= 1138.5 kJ/mol
Therefore, the enthalpy change for the process CCl4(g) -> C(g) is 1138.5 kJ/mol.
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Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. Can you explain this answer?
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Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. 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 Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. 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 Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. Can you explain this answer?.
Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. 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 Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. 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 Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. Can you explain this answer?.
Solutions for Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Calculate the enthalpy change for the process CCI4(g) C (g) → 4 CI (g) and calculate bond enthalpy of C-- CI in CCl4(g) Δvap, H° (CCl4) = 30.5 kJ mol-1 ΔfH0(CCl4) = — 135.5 kJ mol-1 ΔaH° (C) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation ΔaH°(Cl2)= 242 kJ mol-1a)397 kJmol−1b)345 kJmol−1c)307 kJmol−1d)327 kJmol−1Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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