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

i really do not know how to solve this i got no answere

Thanx

 

Answer is not cming...

298 x 8.3 x -2= -4946. -5133 - 4946= 10079. I don't know how it is being done.

Its ok Thanks for ans

This is correct

thank you

Please find this answer

199(493?748+7558=?

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

Please find this answer

Wah.

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