Answer : The correct option is, (B) -5448 kJ/mol
Explanation :
First we have to calculate the heat required by water.
![q=m\times c\times (T_2-T_1)](https://tex.z-dn.net/?f=q%3Dm%5Ctimes%20c%5Ctimes%20%28T_2-T_1%29)
where,
q = heat required by water = ?
m = mass of water = 250 g
c = specific heat capacity of water = ![4.18J/g.K](https://tex.z-dn.net/?f=4.18J%2Fg.K)
= initial temperature of water = 293.0 K
= final temperature of water = 371.2 K
Now put all the given values in the above formula, we get:
![q=250g\times 4.18J/g.K\times (371.2-293.0)K](https://tex.z-dn.net/?f=q%3D250g%5Ctimes%204.18J%2Fg.K%5Ctimes%20%28371.2-293.0%29K)
![q=81719J](https://tex.z-dn.net/?f=q%3D81719J)
Now we have to calculate the enthalpy of combustion of octane.
![\Delta H=\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7Bq%7D%7Bn%7D)
where,
= 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:
![\Delta H=\frac{-81719J}{0.015mole}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-81719J%7D%7B0.015mole%7D)
![\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%3D-5447933.333J%2Fmol%3D-5447.9kJ%2Fmol%5Capprox%20-5448kJ%2Fmol)
Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.