Question 14.9: Equilibrium Constant Evaluate ln K at 298 K for the followin...
Equilibrium Constant
Evaluate ln K at 298 K for the following elementary steps: (a) H _{2}+0.5 O _{2} \rightarrow H _{2} O and (b) 2 H _{2} O \rightarrow 2 H _{2}+ O _{2} . Assume water to be in the vapor phase. What-if scenario: What would the answer in part (a) be if the temperature was (c) 1000 K or (d) 2500 K?
Evaluate \Delta \bar{g}_{T}^{ o } for an elementary step using properties from the ideal gas tables or the n-IG state TESTcalc, then obtain ln K from Eq. (14.49).
\ln K \equiv-\frac{\Delta \bar{g}_{T}^{ o }}{\bar{R} T} \quad \text { or } \quad K \equiv \exp \left(-\frac{\Delta \bar{g}_{T}^{ o }}{\bar{R} T}\right) (14.49)
Assumptions
The IG mixture model is applicable.
Analysis
\Delta \bar{g}_{T}^{ o } can be expressed in terms of enthalpy and entropy as:
\Delta \bar{g}_{T}^{ o }=\Delta\left[\bar{h}\left(p_{0}, T\right)-T \bar{s}\left(p_{0}, T\right)\right]=\Delta \bar{h}(T)-T \Delta \bar{s}^{ o }(T)
Using the procedure developed in the last chapter, we evaluate \Delta \bar{h}(T) \text { and } \Delta \bar{s}^{ o }(T) for the elementary step H _{2}+0.5 O _{2} \rightarrow H _{2} O .
\begin{aligned}\Delta \bar{h}(T) &=\bar{h}_{P}(T)-\bar{h}_{R}(T)=\sum_{p} \nu_{p} \bar{h}_{p}(T)-\sum_{r} \nu_{r} \bar{h}_{r}(T) \\&=\nu_{ H _{2} O } \bar{h}_{f, H _{2} O }^{ o }-\nu_{ H _{2}} \bar{h}_{f, H _{2}}^{ o }-\nu_{ O _{2}} \bar{h}_{f, O _{2}}^{ o } \\&=(1)(-241,820)-(1)(0)-(0.5)(0)=-241,820 kJ / kmol\end{aligned}
\begin{aligned}\Delta \bar{s}^{ o }(T) &=\bar{s}_{P}^{ o }(T)-\bar{s}_{R}^{ o }(T)=\sum_{p} \nu_{p} \bar{s}_{p}^{ o }(T)-\sum_{r} \nu_{r} \bar{s}_{r}^{ o }(T) \\&=\nu_{ H _{2} O } \overline{ S }_{ H _{2}O @ 298 K }^{ o } -\nu_{ H _{2}} \bar{S}_{ H _{2}@ 298 K}^{ o } -\nu_{ O _{2}} \bar{S}_{ O _{2}@ 298 K}^{o} \\&=(1)(188.7)-(1)(130.7)-(0.5)(205.0) \\&=-44.5 kJ / kmol \cdot K\end{aligned}
Combining,
\begin{aligned}\Delta \bar{g}_{T}^{ o } &=\Delta \bar{h}(T)-T \Delta \bar{s}^{ o }(T) \\&=-241,820-(298)(-44.5)=-228,559 kJ / kmol\end{aligned}
Therefore,
\ln K \equiv-\frac{\Delta \bar{g}_{T}^{ o }}{\bar{R} T}=-\frac{-228,559}{(8.314)(298)}=92.25
The second reaction, 2 H _{2} O \rightarrow 2 H _{2}+ O _{2} , represented here by the suffix (b), is simply the inverse of the first reaction multiplied by 2. Therefore,
\Delta \bar{g}_{T(b)}^{o}=-(2)(-228,559)=(2)(228,559) kJ / kmol
and \ln K_{(b)} \equiv-\frac{\Delta \bar{g}_{T(b)}^{ o }}{\bar{R} T}=-\frac{(2)(228,559)}{(8.314)(298)}=-(2)(92.25)=-184.5
TEST Analysis
Launch the IGE system-state TESTcalc. In the composition panel, select the reactants, H2 and O2 for the first reaction and enter their stoichiometric coefficients, 1 and 0.5, in kmol (to register the change you must click on an empty cell). Select H2O in the products block. In the state panel, enter p1=100 kPa and T1=298 K, select the Products radio button, and click Calculate. Obtain ln K from the composition panel as 92.26.
What-if scenario
Change T1 to 1000 K and re-calculate State-1. Evaluate ln K as 23.16. At 2500 K ln K can be computed as 5.13.
Discussion
The huge value of the equilibrium constant e^{92.26} at 298 K, indicates that the forward reaction for the elementary step H _{2}+0.5 O _{2} \rightarrow H _{2} O dominates at room temperature; and hydrogen and oxygen are almost nonexistent at equilibrium (that is why water is so stable at atmospheric conditions). As the temperature is increased, there is a drastic drop in the value of K. At 2500 K, therefore, we will expect some hydrogen and oxygen to be present in the mixture as the inverse reaction strengthens. We will verify that in the next example.