Question

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371.2 K. What is the enthalpy of combustion of octane? The specific heat capacity of water is 4.18 J/K g.

A. -1226 kJ/mol
B. -5448 kJ/mol
C. 293.25 kJ/mol
D. 1226 kJ/mol

Answer 1

B.) -5448 kJ/mol

Step-by-step explanation:

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Answer 2

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

`q=m* c* (T_2-T_1)`

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water =
`4.18J/g.K`

`T_1` = initial temperature of water = 293.0 K

`T_2` = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

`q=250g* 4.18J/g.K* (371.2-293.0)K`

`q=81719J`

Now we have to calculate the enthalpy of combustion of octane.

`\Delta H=(q)/(n)`

where,

`\Delta H` = 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=(-81719J)/(0.015mole)`

`\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol`

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.