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Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate combustion energies (ΔE) for benzene in both kilojoules per gram and kilojoules per mole.

Respuesta :

Answer:

[tex]-3272[/tex]     kJ/mol

Explanation:

Given and known facts

Mass of Benzene [tex]= 0.187[/tex] grams

Mass of water [tex]= 250[/tex] grams

Standard heat capacity of water [tex]= 4.18[/tex] J/g∙°C

Change in temperature ΔT [tex]= 7.48[/tex]°C

Heat

[tex]=250 * 4.18 * 7.48\\=7816.6 \\=7.82[/tex]

Heat released by benzine is - 7.82 kJ

Now, we know that

[tex]0.187[/tex] grams of benzene release [tex]= -7.82[/tex]  kJ heat

So, [tex]1[/tex] g benzine releases

[tex]\frac{ -7.82 }{0.187}\\= -41.8[/tex]

kJ/g

[tex]0.187 * \frac{1}{78.108}=0.00239[/tex] mol C6H6

Heat released

[tex]= \frac{-7.82}{ 0.00239}[/tex]

[tex]=-3272[/tex]     kJ/mol

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