Question

In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess of oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 $kJ\, {{K}^{-1}}$, the numerical value for the enthalpy of combustion of the gas in $kJ\,mo{{l}^{-1}}$ is:

Answer 1

To find the energy release due to the combustion of 3.5 gm of a gas at constant volume we have a formula = $C\times \Delta T$ and thus the energy released by the combustion of 2.8 gm(1moles) of a gas will be, $C\times \Delta T\times \dfrac{28}{3.5}$, where C is the heat capacity and delta T is the change in temperature.

Complete answer:
From your chemistry lessons you have learned about the calorimeter and the heat capacity.
As we know that measuring thermal energy itself is not easy, but we can measure the change in temperature caused by the flow of thermal energy between the objects.
Calorimeter is an instrument or a device used to measure the enthalpy change in a chemical process and the sets of techniques used to measure enthalpy changes is known as calorimetry.
Heat capacity is the amount of energy required to raise the temperature of the object by 1⁰C.Heat capacity is represented at 'C' and its unit is joule per degree celsius but it is expressed more correctly in SI unit that is joules per Kelvin ($J{{K}^{-1}}$).
As we know that the temperature of any objects change when it loses or absorb thermal energy and the magnitude of change in temperature depends on both the heat capacity (C) of an object and the amount of thermal energy that is transferred (q)
The heat capacity (C) is given as,
$C=q\times \Delta T$
Where C= heat capacity,
q = amount of energy transferred and
$\Delta T$ = Change in temperature.
So, in the question the heat capacity is given as 2.5 $kJ{{K}^{-1}}$ and the change in temperature ($\Delta T$) is 298 K to 298.45 K,
So, energy release due to the combustion of 3.5 gm of a gas at constant volume we have a formula = $C\times \Delta T$= 2.5× (298.45-298) = 2.5× 0.45
Thus the energy released by the combustion of 28 gm (1 mole) of a gas will be
$2.5\times 0.45\times \dfrac{28}{3.5}=9\,kJmo{{l}^{-1}}$

Note: Change in temperature will always be written as final temperature- initial temperature ($\Delta T={{T}_{final}}-{{T}_{initial}}$). The value of heat capacity can never be negative; it is an intrinsically positive number but the value of $\Delta T$ and q can either be negative or positive and the signs of both of them will be the same. If they are positive it means heat is flowing into the object from the surrounding and if negative then heat flows out of the object in the surrounding.