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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.**

*constant volume*

**constant pressure.**

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 :

${C}_{10}{H}_{8}+12\hspace{0.17em}{O}_{2}\to 10C{O}_{2}+\hspace{0.17em}4{H}_{2}O$

So, the change in the number of gaseous molecule $\u2206n=10-12=-2$

And the relation between q

_{p}and q

_{v}is:

${q}_{p}={q}_{v}+\u2206nRT\hspace{0.17em}$

Substituting the values given :

${q}_{p}=-5138.8\times {10}^{3}+(-2\times 8.314\times 298)\phantom{\rule{0ex}{0ex}}=-5138800-4955\phantom{\rule{0ex}{0ex}}=-5143\times {10}^{3}J/mol\phantom{\rule{0ex}{0ex}}=-5143.75KJ/mol$

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

Regards.

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