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If at 298 K the bond energies of C — H, C — C, C = C and H — H bonds are respectively 414, 347, 615 and 435 kJ mol–1, the value of enthalpy change for the reaction H2C = CH2(g) + H2(g) → H3C — CH3(g) at 298 K will be [2003] 
  • a)
    – 250 kJ
  • b)
    + 125 kJ
  • c)
    – 125 kJ
  • d)
    + 250 kJ
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If at 298 K the bond energies of C — H, C — C, C = C and H...
CH2 = CH2(g) + H2(g) → CH3 - CH3
Enthalpy change = Bond energy of reactants – Bond energy of products.
ΔH = 1(C = C) + 4 (C – H) + 1 (H - H) - 1 (C - C) - 6 (C - H)
         = 1 (C = C) + 1 ( H – H) – 1 (C – C) – 2 (C– H)
        = 615 + 435 – 347 – 2 × 414  = 1050 – 1175 = –125 kJ.
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If at 298 K the bond energies of C — H, C — C, C = C and H — H bonds are respectively 414, 347, 615 and 435 kJ mol–1, the value of enthalpy change for the reaction H2C = CH2(g) + H2(g) → H3C — CH3(g) at 298 K will be [2003]a)– 250 kJb)+ 125 kJc)– 125 kJd)+ 250 kJCorrect answer is option 'C'. Can you explain this answer?
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If at 298 K the bond energies of C — H, C — C, C = C and H — H bonds are respectively 414, 347, 615 and 435 kJ mol–1, the value of enthalpy change for the reaction H2C = CH2(g) + H2(g) → H3C — CH3(g) at 298 K will be [2003]a)– 250 kJb)+ 125 kJc)– 125 kJd)+ 250 kJCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If at 298 K the bond energies of C — H, C — C, C = C and H — H bonds are respectively 414, 347, 615 and 435 kJ mol–1, the value of enthalpy change for the reaction H2C = CH2(g) + H2(g) → H3C — CH3(g) at 298 K will be [2003]a)– 250 kJb)+ 125 kJc)– 125 kJd)+ 250 kJCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If at 298 K the bond energies of C — H, C — C, C = C and H — H bonds are respectively 414, 347, 615 and 435 kJ mol–1, the value of enthalpy change for the reaction H2C = CH2(g) + H2(g) → H3C — CH3(g) at 298 K will be [2003]a)– 250 kJb)+ 125 kJc)– 125 kJd)+ 250 kJCorrect answer is option 'C'. Can you explain this answer?.
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