# the heat of combustion of naphthalene at constant volume at 25 degree celsius was found to be -5133 kj/mole calculate the value of enthalpy change at constant pressure at the same temperature

The heat of combustion at constant volume is given to be -5133 KJ/mol.
Therefore, qv=-5133 kJ/mol.

The balanced equation of combustion of naphthalene is:

C10H8 + 12O-----> 10CO2 + 4H​2O

Here, the number of gaseous moles on the products side= 10 (10 moles of CO2)
And, the number of gaseous moles on the reactants side= 12(12 moles of O2)

Hence, the change in number of gaseous moles, Δn = 10 - 12 = -2

The relation between qp and qv is:
qp = qv + ΔnRT

Substituting the values given,
q= -5133*10+ (-2*8.314*298)
=> qp= -5137955.144 J/mol

• 153
i really do not know how to solve this i got no answere
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thank you soooo much
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Thanx

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298 x 8.3 x -2= -4946. -5133 - 4946= 10079. I don't know how it is being done.
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Its ok Thanks for ans
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This is correct • 36
thank you
• 1 • 0
199(493?748+7558=?
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The heat of combustion at constant volume is given to be -5133 KJ/mol.
Therefore, qv=-5133 kJ/mol.

The balanced equation of combustion of naphthalene is:

C10H8 + 12O2 -----> 10CO2 + 4H?2O

Here, the number of gaseous moles on the products side= 10 (10 moles of CO2)
And, the number of gaseous moles on the reactants side= 12(12 moles of O2)

Hence, the change in number of gaseous moles, ?n = 10 - 12 = -2

The relation between qp and qv is:
qp = qv + ?nRT

Substituting the values given,
qp = -5133*103 + (-2*8.314*298)
=> qp= -5137955.144 J/mol
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