Chapter 15, Problem 15.111QP

Calculate ΔG° and K_p for the following processes at 25°C.

$\begin{array}{l}\left(\text{a}\right){\text{H}}_{\text{2}}\left(g\right)+{\text{Br}}_{\text{2}}\left(l\right)\rightleftarrows \text{2HBr}\left(g\right)\\ \left(\text{b}\right)\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\left(g\right)+\frac{\text{1}}{\text{2}}{\text{Br}}_{\text{2}}\left(l\right)\rightleftarrows \text{HBr}\left(g\right)\end{array}$

Account for the differences in ΔG° and K_p obtained for parts (a) and (b).

Interpretation Introduction

Interpretation:

The standard free energy $({\text{ΔG}}^{\text{ο}})$ and equilibrium constant (Kp) values should be calculated given the heterogeneous equilibrium reaction with respective temperature at ${\text{25}}^{\text{0}}\text{C}$.

Concept Introduction:

Chemical equilibrium: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

Entropy $({\text{ΔG}}^0)$: The entropy is second law of thermodynamics it indicated for $({\text{ΔG}}^0)$ symbol, the many of chemical reactions cause changes in entropy and it plays on important role in determining in which direction (forward and backward) a chemical reaction spontaneously proceeds.

Gibbs free energy (G): The thermodynamic quantity to the ( $\text{ΔH}$) enthalpy of a system process and minutes the product of entropy and the absolute temperature. The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system.

Heterogeneous equilibrium: This equilibrium reaction does not depend on the amounts of pure solid and liquid present, in other words heterogeneous equilibrium, substances are in different phases.

Answer to Problem 15.111QP

The equilibrium pressure (Kp) value of given the homogenies reaction is showed below.

$\begin{array}{l}(a).H_2(g)+B{r}_2(l)\rightleftharpoons 2HBr(g)\\ \Delta G^{\text{ο}}=-106.4\text{KJ/mol}\\ {\text{K}}_{\text{P}}=3.87\times {10}^{18}\\ (\text{b)}\text{.}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{Br}}_{\text{2}}\text{(l)}\rightleftharpoons \text{HBr}\text{(g)}\\ \Delta G^{\text{ο}}=-53.2\text{KJ/mol}\\ {\text{K}}_{\text{P}}=2.17\times {10}^9\end{array}$

Explanation of Solution

To find: Calculate the standard entropy $({\text{ΔG}}^0)$ and partial pressure (Kp) values for given the statement (a).

 (a)

Calculate the heterogeneous equilibrium process $({\text{ΔG}}^0)$ and (Kp) at 25⁰C.

We consider the following standard free energy equation

$\begin{array}{l}(a).H_2(g)+B{r}_2(l)\rightleftharpoons 2HBr(g)\\ {\text{ΔG}}^{\text{0}}={\text{ΔG}}_f^0\text{(Products)-}{\text{ΔG}}_f^o(Reac\tan ts)\\ (or){\text{ΔG}}_{\text{rxn}}^0=\sum n{\text{ΔG}}_{\text{f}}{}^{\text{0}}\text{(Products)-}\sum m{\text{Δ}}_{\text{f}}{}^0(Reac\tan ts)\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=2{\text{ΔG}}_{\text{f}}^0\text{(HBr)}-\text{Δ}{\text{G}}_{\text{f}}^0{\text{(H}}_{\text{2}})-\text{Δ}{\text{G}}_{\text{f}}^0{\text{(Br}}_{\text{2}})\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=2(-53.2KJ/mol)-(1)(0)-(1)(0)(\because \text{The respactive}{\text{ΔG}}_{\text{f}}^{0}KJ/molvaluestakingfromequilibriumtable)\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=-106.4\text{KJ/mol}\\ \end{array}$ Calculate the partial pressure values (Kp)

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K---------[1]}\\ \text{ΔG}=\text{Free energy}\\ {\text{ΔG}}^{\text{0}}=\text{Standard state free energy}\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\\ \ln =(-ve(\log )\text{State Function})\\ \text{Let}\text{us consider following equation (1)}\\ {\text{ΔG}}^{\text{0}}=-106.4\text{KJ/mol}\text{(216}\text{.1}\text{KJ/mol}(1K={10}^3)\text{converted into a J/mol)}\\ \Delta {\text{G}}^{\text{0}}\text{values substituted equation (1) and rewrite the above equation}\\ \text{InK}\text{p=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{\text{-(106}{\text{.4×10}}^3\text{J/mol)}}{\text{(8}\text{.314}\text{J/K}\text{mol)(298}\text{K)}}\\ \text{InK}{}_{\text{P}}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{-106.4{\text{×10}}^3\text{J/mol}}{(2477.572)}=42.9(Here\text{R=}\text{8}\text{.314,}{\text{T=273+25}}^{\text{0}}\text{C=298K)}\\ \text{Kp=}{e}^{42.9}(Herelo{g}_e=1)\\ \text{Kp=}3.87\times {10}^{18}\end{array}$

The standard free energy values and partial pressure values are derived given the equilibrium statement (a) and its temperature.

To find: Calculate the standard entropy $({\text{ΔG}}^0)$ and partial pressure (Kp) values for given the statement (b).

Calculate the heterogeneous equilibrium process $({\text{ΔG}}^0)$ and (Kp) at 25⁰C.

$\begin{array}{l}(\text{b)}\text{.}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{Br}}_{\text{2}}\text{(l)}\rightleftharpoons \text{HBr}\text{(g)}\\ {\text{ΔG}}^{\text{0}}={\text{ΔG}}_f^0\text{(Products)-}{\text{ΔG}}_f^o(Reac\tan ts)\\ (or){\text{ΔG}}_{\text{rxn}}^0=\sum n{\text{ΔG}}_{\text{f}}{}^{\text{0}}\text{(Products)-}\sum m{\text{Δ}}_{\text{f}}{}^0(Reac\tan ts)\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}={\text{ΔG}}_{\text{f}}^0\text{(HBr)}-\frac{1}{2}\text{Δ}{\text{G}}_{\text{f}}^0{\text{(H}}_{\text{2}})-\frac{1}{2}\text{Δ}{\text{G}}_{\text{f}}^0{\text{(Br}}_{\text{2}})\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=1(-53.2KJ/mol)-(\frac{1}{2})(0)-(\frac{1}{2})(0)(\because \text{The respactive}{\text{ΔG}}_{\text{f}}^{0}KJ/molvaluestakingfromequilibriumtable)\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=-53.2\text{KJ/mol}\end{array}$

Calculate the partial pressure values (Kp)

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K---------[1]}\\ \text{Let}\text{us consider following equation (1)}\\ {\text{ΔG}}^{\text{0}}=-53.2\text{KJ/mol}\text{(-53}\text{.2}\text{KJ/mol}(1K={10}^3)\text{converted into a J/mol)}\\ \Delta {\text{G}}^{\text{0}}\text{values substituted equation (1) and rewrite the above equation}\\ \text{InK}\text{p=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{\text{-(53}{\text{.2×10}}^3\text{J/mol)}}{\text{(8}\text{.314}\text{J/K}\text{mol)(298}\text{K)}}\\ \text{InK}{}_{\text{P}}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{-53.2{\text{×10}}^3\text{J/mol}}{(2477.572)}=21.5(Here\text{R=}\text{8}\text{.314,}{\text{T=273+25}}^{\text{0}}\text{C=298K)}\\ \text{Kp=}{e}^{21.5}(Herelo{g}_e=1)\\ \text{Kp=}2.17\times {10}^9\end{array}$

The given decomposition reaction the respective reactant to give products all exists in the different phase  and this equilibrium reaction expression contains single conditions like gases phase, the equilibrium constant can also be represented by Kp and Kc, were the “P” partial pressure. The Kp derived equation showed above.

Conclusion

The equilibrium pressure (Kp) values are derived given both statement of homogenius equilibrium reactions.