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.)

Short 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.

Step-by-step solution

Given Information : 

One mole of H₂

1/2 mole of O₂

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 $PV$on the system. The enthalpy change is the total heat emitted by the system as a result of these two mechanisms.

Explanation 

The energy resulting from the $PV$ 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:

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

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.