Very Very Very urgent please help
One mole of napthalene was burnt in oxygen gas at  constant volume  to give CO2 gas and water at 25 C. The heat evolved was found to be 5138.8kJ.Calculate the heat of reaction at constant pressure.
 

Dear Student,

The heat of combustion at constant volume is given to be - 5138.8 kJ /mol.
Therefore q_v = - 5138.8 ​kJ /mol.

The balanced equation of the combustion of naphthalene is : 

$C_{10}{H}_8 +12\hspace{0.17em}{O}_2 \to 10C{O}_2 +\hspace{0.17em}4H_{2}O$

So, the change in the number of gaseous molecule $∆n = 10 - 12 = -2$

And the relation between q_p and q_v is: 

$q_p=q_v+ ∆nRT\hspace{0.17em}$

Substituting the values given : 

$$\begin{gathered} q_{p }= - 5138.8 \times {10}^3 + (-2\times 8.314 \times 298) \\ = - 5138800 - 4955 \\ = -5143 \times {10}^3 J/ mol \\ = - 5143.75 KJ/mol \end{gathered}$$

So the heat of reacion at constant pressure is -5143.75 KJ/mol

Regards.