# Is enthalpy of combustion the 'heat released' or 'enthalpy change'?

In a constant volume calorimeter, $$\pu{3.5 g}$$ of a gas with molar weight $$\pu{28 g}$$ was burnt in excess $$\ce{O2}$$ at $$\pu{298 K}.$$ Temperature of calorimeter increased from $$298$$ to $$\pu{298.45 K}$$ due to combustion. If heat capacity of calorimeter is $$\pu{2.5 J/K},$$ then magnitude enthalpy of combustion of gas in $$\pu{kJ/mol}$$ is?

In this question, according to me $$ΔU = 2.5 × 0.45 = 1.125,$$ i.e. $$1.125/(3.5/28) = \pu{9 kJ/mol}$$, and $$ΔH = ΔU + nRΔT = 1.59.$$

But according to my teacher, we have to find the heat evolved (which in isobaric process equals ΔU). So answer is $$9$$.

This is a question from IIT-JEE so I think there is no error in question or answer.

Isn't enthalpy of combustion the change in enthalpy?

• As a general note, never omit units (also during intermediate calculations); also, molar weight is measured in $\pu{g mol-1},$ not just $\pu{g}$. – andselisk Oct 15 '19 at 14:50

The enthalpy of combustion (aka heat of combustion) is defined as the enthalpy change in going from pure reactants (each at 1 bar pressure) to pure products (again, each at 1 bar pressure) at 298 K.

In this problem, it is first of all necessary to assume ideal gas behavior to proceed. For ideal gases, the change in enthalpy in going from pure reactants each at 1 bar pressure to the starting gas mixture in the calorimeter, there is no change in enthalpy at constant temperature. Then, application of the first law of thermodynamics to the mixture before and after the reaction leads to the conclusion that $$\Delta U=Q$$, since the reaction volume is constant. Since the temperature change between the beginning and end of the combustion in the calorimeter is essentially zero, this means the the calorimeter has measured the internal energy change of the reaction. As Buck Thorn has pointed out, the change in enthalpy for the gases in the calorimeter is given for an ideal gas by: $$\Delta H=\Delta U+\Delta (PV)=\Delta U+RT\Delta n$$where $$\Delta n$$ is the stoichiometric change in the number of moles of gas. He also correctly pointed out that there is nothing said in the problem statement about the stoichometry of the reaction or the change in the number of moles of gas. So, as he indicated, the problem statement is flawed. However, if we arbitrarily assume that the change in the number of moles of gas is zero, we would have to conclude that the change in enthalpy for the process in the calorimeter is equal to the change in internal energy. Then, since, again, the products form an ideal gas mixture, there is no change in enthalpy in going from the final gas mixture in the calorimeter to the pure products at 1 bar pressure. So, based on these assumptions, the enthalpy of combustion is equal to the internal energy change in the calorimeter (also Q) per mole of species being burned.

Enthalpy of combustion is the change in enthalpy. The enthalpy change of a combustion reaction is called ΔH (combustion). The heat supplied to the calorimeter represents the heat supplied by virtue of the reaction taking place.

Normally, ΔH = ΔU + PΔv

The calorimeter is constant volume, so ΔV = 0. Hence, ΔH = ΔU in this case. So, as per the given conditions, for this particular reaction, the enthalpy of combustion and the internal energy change of combustion are the same.

So, usually when a problem like this is given, the heat absorbed by the calorimeter is ΔH, but here ΔH = ΔU given that no volume change occurs.

So, normally what we do is, we write ΔH = ΔU + PΔV, right? And PΔV = nRΔT. But this is for the gas! What you have done is you have taken the calorimeter to be an ideal gas! Hence ΔH = ΔU + nRΔT is applicable to the ΔT of the gas, not to the ΔT of the calorimeter. (On a side note, you are making an ideal gas assumption by assuming PΔV = nRΔT to be valid, which we cannot be sure of).

To summarise,

-> Heat absorbed by calorimeter represents the enthalpy change of the reaction

-> ΔH = ΔU + PΔV. ΔV = 0, so ΔH = ΔU

-> PΔV = nRΔT is applicable to ideal gases only! Do NOT take the ΔT of the calorimeter as we don't have any data about temperature changes of the gases inside.

-> I think for most IIT-JEE problems, the ideal gas assumption is valid. But use PΔV first, don't jump to nRΔT without thinking.

-> Think about which ΔT you're using and whose temperature is changing.

• H = U + PV So, ΔH = ΔU + Δ(PV) = ΔU + PΔV + VΔP = ΔU + VΔP Since volume is constant but pressure is not. I did not understand why you assumed pressure to be constant. – Shashwat Oct 15 '19 at 15:25

For an isobaric process $$q_p=\Delta H$$ and, as you rightly write, for an isochoric process $$q_V=\Delta U$$ and the relation between $$U$$ and $$H$$ is \begin{align}\Delta H&=\Delta U + \Delta(PV)\\&=\Delta U+RT\Delta n\end{align} The last equality holds only if T is constant, which it evidently wasn't during the experiment, but for our calculation of $$\Delta H$$ we are allowed to assume this since we are interested in the enthalpy change for the system (which comprises the reaction, not the calorimeter) at constant T, and we can safely assume that $$\Delta U$$ and $$\Delta H$$ of the system are approximately independent of temperature over the small range of T in the experiment, but possibly dependent on n. The same assumptions cannot be made for the calorimeter: its energy is assumed here to depend entirely on T.

The next step is figuring out $$\Delta n$$. The only way you can have $$\Delta H=\Delta U$$ for an isothermal ideal gas is if you end up with the same molar amount of gas you started with ($$\Delta n=0$$), which is necessary based on the stated solution but seems ungrounded. A hint might be the MW, which suggests the gas is perhaps $$\ce{N2}$$ or ethene, or even carbon monoxide, but none of these combust with the desired change in stoichiometry (well, maybe ethene, if you assume the water produced does not condense). So I'd charge that the question is flawed.