Calculating internal energy change of a reaction

Question

We are asked to find the relation between $\Delta H$ and $\Delta U$, and also, the sign of the internal energy change of reaction: $$\ce{CO(g) +\dfrac{1}{2}O2(g) -> CO2(g)}$$ This was part (3) of a question, and this one explicitly mentioned: "reaction happens in an isolated container". The other parts had "closed container".

Approach: For ideal gases, we know that $U=nCvT$. So, we have:

$U_{rectants}:(1*5R/2*T) +(1/2*5R/2*T)=\dfrac{15}{4}RT$

$U_{products}:(1*3R*T)=3RT$.

Thus $\Delta U=U_{products}-U_{rectants}=\dfrac{3}{4}RT>0.$

However, the answer given is $\Delta U$ is $0$.

And the answer also claims that $\Delta H$<$\Delta U$ , which is expected from the relationship $\Delta H$=$\Delta U$+$RT\Delta n$ is not true.

Now I do not know exactly how do we account for "isolated system" in the formula for internal energy.

What exactly does isolated system refer to in this context? And how exactly is it affecting the internal energy?

A possible approach might have to do with the FLOT. I (guess) that isolated system means that there is no heat exchange with the surroundings, so the expression reduces to $dU+P_{ext}dV=0$. Does "isolated system" somehow also imply that $P_{ext}dV=0$?

Answer 1

An isolated system is one that can exchange no mass, heat, or work with the surroundings. So, from the first law, the change in internal energy of an isolated system in any process within the system is $\Delta U=0$.

Now for my analysis of this problem in greater detail to determine the enthalpy change. I'm taking as the initial state of the system one mole of CO and 1/2 mole of O2 at 298 K and 1 bar in a rigid adiabatic container, and the final state as 1 mole of reaction product CO2 in this same container at whatever temperature and pressure result when the reaction occurs spontaneously in this same adiabatic rigid container.

My game plan is to examine the internal energy change for the process implemented in 3 successive steps, based on Hess' law to determine the overall internal energy change:

1. Reactants reacting at constant pressure and removing whatever heat is necessary to hold the temperature constant at 298 K. So the final state in step 1 is one mole of CO2 product at 298 K and 1 bar, at 2/3 the original volume (since 3/2 moles goes to 1 mole).

2. Expanding the 1 mole of CO2 isothermally from 2/3 the original volume back up to the full original volume, while decreasing the pressure from 1 bar to 2/3 bar. So the final state of step 2 is 1 mole of CO2 at 298 K and 2/3 bar pressure.

3. Adding heat to the 1 mole of CO2 at constant volume to increase its temperature to whatever value is necessary for the overall change in the internal energy for combination of the 3 steps to be equal to zero.

Analysis of internal energy change for step 1:

The process described for step 1 corresponds to the conditions for the standard heat of the reaction CO + 1/2 O2--> CO2 at 298 K and 1 bar, $\Delta H^0=-283000\ Joules$. So the change in internal energy for this step is the standard change in internal energy $$\Delta U^0=\Delta H^0-RT^0\Delta n=\Delta H^0+\frac{RT^0}{2}=-283000-(8.314)(298)(-1/2)=-282000\ Joules$$

Analysis of internal energy change for step 2:

Step 2 involves an isothermal change for an ideal gas (without chemical reaction), so the internal energy change in step 2 for the system in this step is zero.

Analysis off internal energy change for step 3:

Step 3 involves the constant volume heating of one mole of the ideal gas O2 from $T^0=298 K$ to the final temperature T, so the change in internal energy from this step is $$(\Delta U)_3=C_v(T-T^0)$$ where $C_v$ is the molar heat capacity of CO2.

Overall combined internal energy change

It follows from all this that the overall internal energy change for this combined 3 step process is $$\Delta U=\Delta H^0+\frac{RT^0}{2}+C_v(T-T^0)$$But we know that this overall internal energy change must be equal to zero. This enables us to calculate the temperature rise of the system in the adiabatic rigid container: $$(T-T^0)=-\frac{\left(\Delta H^0+\frac{RT^0}{2}\right)}{C_v}$$The corresponding enthalpy change for the combined process is $$\Delta H=\Delta H^0+C_p(T-T^0)$$After some algebraic manipulation, this reduces to $$\Delta H=-(\gamma-1)\left(\Delta H^0+\frac{C_pT^0}{2}\right)$$Substituting literature values for the molar heat capacities and heat capacity ratio into this equation then yields the enthalpy change for the overall process: $$\Delta H=+83000\ Joules$$ This indicates that answers B and D are both correct.

So, for this problem, we find that $$\Delta H>\Delta U=0$$

The main error you made in analyzing this (difficult) problem was failing to account for the heat of reaction (where the internal energy and enthalpy can change at constant temperature).