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The enthalpy of combustion of C(graphite ) and C (diamond) are -393.5 and -395.4 KJ mol⁻1 respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol⁻1 is
  • a)
    -1.9
  • b)
    -788.9
  • c)
    1.9
  • d)
    788.9
Correct answer is option 'C'. Can you explain this answer?
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The enthalpy of combustion of C(graphite ) and C (diamond) are -393.5 ...
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The enthalpy of combustion of C(graphite ) and C (diamond) are -393.5 ...
Enthalpy of combustion of C(graphite) = -393.5 KJ/mol
Enthalpy of combustion of C(diamond) = -395.4 KJ/mol

Enthalpy of combustion is the heat released when one mole of a substance is burnt completely in the presence of oxygen.

Enthalpy of combustion of C(graphite) is less than that of C(diamond) which means that C(graphite) is more stable than C(diamond).

Enthalpy of conversion of C(graphite) to C(diamond) can be calculated using the following formula:

ΔH = Enthalpy of products - Enthalpy of reactants

Here, the reactant is C(graphite) and the product is C(diamond).

ΔH = (-395.4 KJ/mol) - (-393.5 KJ/mol)
ΔH = -1.9 KJ/mol

The enthalpy of conversion of C(graphite) to C(diamond) is -1.9 KJ/mol. Since the question is asking for the answer in KJ/mol, the correct option is C) 1.9 KJ/mol.
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The enthalpy of combustion of C(graphite ) and C (diamond) are -393.5 and -395.4 KJ mol⁻1 respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol⁻1 isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer?
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