Question

When 2.00 g of methane are burned in a bomb calorimeter, the change in temperature is 3.08°C. The heat capacity of the calorimeter is 2.68 kJ/°C. The molar mass of methane is 16.042 g/mol. What is the approximate molar enthalpy of combustion of this substance?

Answer 1

Answer:

The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

$q=c\times \Delta T$

where,

q = Heat gained = ?

c = Specific heat = $2.68 kJ/^oC$

ΔT =  The change in temperature = 3.08°C

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

$q=2.68 kJ/^oC\times 3.08^oC$

$q=8.2544 kJ$

Now we have to calculate molar enthalpy of combustion of this substance :

$\Delta H_{comb}=-\frac{q}{n}$

where,

$\Delta H_{comb}$ = enthalpy change = ?

q = heat gained = 8.2544kJ

n = number of moles methane = $\frac{\text{Mass of methane}}{\text{Molar mass of methane }}=\frac{2.00 g}{16.042 g/mol}=0.1247 mole$

$\Delta H_{comb}=-\frac{8.2544 kJ}{0.1247 mole}=-66.21 kJ/mole\approx -66 kJ/mole$

Therefore,  the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.