# Q. In a constant volume calorimeter,3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298 K.The temperature of the calorimeter was found to increase from 298 to 298.45 K due to...

Q. In a constant volume calorimeter,3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298 K.The temperature of the calorimeter was found to increase from 298 to 298.45 K due to the combustion process.Given that heat capacity of the calorimeter is 2.5 `kJ K^-1` .The numerical value for the enthalpy of combustion of gas in `kJ mol^-1` is:-

mvcdc | Student, Graduate | (Level 2) Associate Educator

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We know the heat capacity of the calorimeter. The heat capacity gives us the amount of heated needed to be absorbed in order to raise the temperature of a material. In this case, the calorimeter has a heat capacity of `2.5(kJ)/K` . This means that a `1K` increase in heat will be the effect by an input of `2.5kJ` of heat. ``

The temperature was increased from `298K`  to`298.45K` . Hence, `Delta T = 298.45K - 298K = 0.45K.`

The total heat absorbed then is `Q = CDeltaT = 2.5(kJ)/K times 0.45K = 1.125 kJ.`

This amount of heat is due to the combustion of a total of `3.5 g` of gas. To get the enthalpy change in `(kJ)/(mol)` , we simply get the number of moles of the gas:

`n = (3.5g)/(28(g/(mol))) = 0.125 mol`

Then,

`Delta H = Q/n = (1.125kJ)/(0.125mol) = +9 (kJ)/(mol)`

Where `DeltaH` is the change in enthalpy (molar).

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