Q.1.49 - An Introduction to Thermal Physics

Question

Problem 1.49. Consider the combustion of one mole of ${\mathrm{H}}_2$ with$1 / 2$  mole of ${\mathrm{O}}_2$ under standard conditions, as discussed in the text. How much of the heat energy produced comes from a decrease in the internal energy of the system, and how much comes from work done by the collapsing atmosphere? (Treat the volume of the liquid water as negligible.)

Step-by-Step Solution

Answer

Short Answer :

The quantity of heat produced by the collapsing atmosphere accounts for 1.31 percent of the total, with the remaining 98.69 percent coming from the system's increased internal energy.

Given Information :

One mole of H₂

1/2  mole of O₂ 

Step 1: Explanation

The enthalpy change for  reaction where one mole of hydrogen molecules combines with half a mole of oxygen molecules to produce water is $\Delta H=-2.86\times {10}^5 \mathrm{J}$, assuming that  reactant gases and the resulting water are both at $25°\mathrm{C}$ and $1 \mathrm{atm}$ pressure. Because the water will be in the form of vapour at first, it will need to release heat in order to condense into a liquid and then cool to room temperature. This results in a decrease in the thermal energy U of the system because U depends on the temperature difference, $T_f<T_i$. As well, the atmosphere will fill in the volume originally occupied by the reactant gases, doing work $P V$ on the system. The enthalpy change is the total heat emitted by the system as a result of these two mechanisms.

Step 2 : Explanation

The energy resulting from the $P V$ work is (assuming that the volume of the liquid water is negligible compared to the initial volume) is:

$W=P\Delta V=P\left(V_f-V_i\right)=-P{V}_i$

the final volume  volume of the water and we neglect it $V_f=0$, since it is very small, the reactant gases and the resulting water are both at $25°\mathrm{C}$ and $1\mathrm{atm}$ pressure, so the work (from ideal gas law) is therefore:

$W=-P{V}_i=-nRT$

where $n=0.5 \mathrm{mol}\left(H\right)+1 \mathrm{mol}\left(O\right)=1.5 \mathrm{mol},R=8.31 \mathrm{J}·{\mathrm{K}}^{-1}·{mol}^{-1}$ and $T=25°\mathrm{C}=298°\mathrm{K}$, so:

$W=-1.5\times 8.31\times 298=-3.7\times {10}^3 \mathrm{J}$

Step 3 : Explanation

The change in enthalpy in the reaction is given by:

$$\begin{gathered} \Delta H=\Delta U+W \\ \to \Delta U=\Delta H-W \end{gathered}$$

Substitute, so:

$\Delta U=\left(-2.86\times {10}^5\right)+\left(3.7\times {10}^3\right)=-2.823\times {10}^5 \mathrm{J}$

The work, $P\Delta V$, contribution is:

$$\begin{gathered} \frac{W}{\Delta U}=\frac{-3.7\times {10}^3}{-2.823\times {10}^5}=0.0131 \\ \frac{W}{\Delta U}=1.31\% \end{gathered}$$

The contribution of amount of heat come from the work done by the collapsing atmosphere is $1.31\%$, where the rest for $98.69\%$ comes from the increase of internal energy of the system.