Respuesta :
Answer : The correct option is, (B) -5448 kJ/mol
Explanation :
First we have to calculate the heat required by water.
[tex]q=m\times c\times (T_2-T_1)[/tex]
where,
q = heat required by water = ?
m = mass of water = 250 g
c = specific heat capacity of water = [tex]4.18J/g.K[/tex]
[tex]T_1[/tex] = initial temperature of water = 293.0 K
[tex]T_2[/tex] = final temperature of water = 371.2 K
Now put all the given values in the above formula, we get:
[tex]q=250g\times 4.18J/g.K\times (371.2-293.0)K[/tex]
[tex]q=81719J[/tex]
Now we have to calculate the enthalpy of combustion of octane.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy of combustion of octane = ?
q = heat released = -81719 J
n = moles of octane = 0.015 moles
Now put all the given values in the above formula, we get:
[tex]\Delta H=\frac{-81719J}{0.015mole}[/tex]
[tex]\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol[/tex]
Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.
