Answer:
[tex]\Delta_{r}U[/tex] of the reaction is -6313 kJ/mol
[tex]\Delta_{r}H[/tex] of the reaction is -6312 kJ/mol
Explanation:
[tex]Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C[/tex]
[tex]q_{cal}= C \times \Delta T[/tex]
[tex]=5.861 \times 3.596 = 21.076\,kJ[/tex]
[tex]q_{rxn}= -q_{cal}= -21.076\,kJ[/tex]
[tex]\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol[/tex]
Therefore, [tex]\Delta_{r}U[/tex] of the reaction is -6313 kJ/mol.
The chemical reaction in bomb calorimeter is as follows.
[tex]C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)[/tex]
[tex]Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5[/tex]
[tex]\Delta_{r}H=\Delta E+ \Delta n. RT[/tex]
[tex]=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol[/tex]
Therefore, [tex]\Delta_{r}H[/tex] of the reaction is -6312 kJ/mol.
