Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reaction. 2 NOCl(g) -> 2 NO(g) + Cl2(g) Round your answer to 2 significant digits. K =
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We can do this by using the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. Show more…
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Calculate the value of the thermodynamic equilibrium constant for the following reactions at $25^{\circ} \mathrm{C}$. (Refer to the data in Appendix C.) $\Delta G_{\mathrm{f}, \mathrm{H}_{2} \mathrm{O}_{2}}^{\circ}=-105.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ (a) $\mathrm{N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ (b) $\mathrm{N}_{2} \mathrm{H}_{4}(g)+6 \mathrm{H}_{2} \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(g)$
Calculate the equilibrium constant at $25^{\circ} \mathrm{C}$ for each of the following reactions from the value of $\Delta G^{\circ}$ given. (a) $\mathrm{I}_{2}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{ICl}(g) \quad \Delta G^{\circ}=-10.88 \mathrm{kJ}$ (b) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \rightarrow 2 \mathrm{HI}(g) \quad \Delta G^{\circ}=3.4 \mathrm{kJ}$ (c) $\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g) \quad \Delta G^{\circ}=-39 \mathrm{kJ}$ (d) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \quad \Delta G^{\circ}=-141.82 \mathrm{kJ}$ (e) $\mathrm{CS}_{2}(g) \rightarrow \mathrm{CS}_{2}(l) \qquad \Delta G^{\circ}=-1.88 \mathrm{kJ}$
The equilibrium constant for the reaction given below is $2.0 \times 10^{-7}$ at $300 \mathrm{~K}$. Calculate the standard free energy change for the reaction; $$ \mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} $$ Also, calculate the standard entropy change if $\Delta H^{\circ}=28.40 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
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