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

The approximate molar enthalpy of combustion of methane is 132.597 kJ/mol.

Step-by-step explanation:

To calculate the molar enthalpy of combustion of methane, we need to use the formula:

q = mcΔT

where q is the heat released, m is the mass of the substance, c is the heat capacity of the calorimeter, and ΔT is the change in temperature.

First, we need to calculate the heat released:

q = mcΔT = (2.00 g)(2.68 kJ/°C)(3.08°C) = 16.5912 kJ

Next, we need to convert the mass of methane to moles:

moles of methane = mass of methane / molar mass of methane = 2.00 g / 16.042 g/mol = 0.12482 mol

Finally, we can calculate the molar enthalpy of combustion:

molar enthalpy of combustion = q / moles of methane = 16.5912 kJ / 0.12482 mol ≈ 132.597 kJ/mol

Answer 2

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

Step-by-step explanation:

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

`q=c* \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* 3.08^oC`

`q=8.2544 kJ`

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

`\Delta H_(comb)=-(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 }}=(2.00 g)/(16.042 g/mol)=0.1247 mole`

`\Delta H_(comb)=-(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.