Question

Question:

(a)Calculate the standard free-energy change and the equilibrium constant for the dimerization of${\mathrm{NO}}_2$to ${\mathrm{N}}_2{\mathrm{O}}_4$at $25°\mathrm{C}$ (see Appendix D).

(b)Calculate $\mathrm{\Delta G}$for this reaction at $25°\mathrm{C}$when the pressures of ${\mathrm{NO}}_2$and ${\mathrm{N}}_2{\mathrm{O}}_4$are each held at 0.010 atm. Which way will the reaction tend to proceed?

Short answer

a) The standard free-energy change is -4.73KJ and the equilibrium constant is 6.75 for the dimerization

b) In backward direction the reaction tend to proceed.

Step-by-step solution

Step-by-step solution

a)

At $25°\mathrm{C}$the dimerization reaction as follows,

$2{\mathrm{NO}}_2\left(\mathrm{g}\right)\rightleftarrows {\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)$

By using standard free energy formula and using the values of Appendix D, we get the value as follows,

localid="1660279889554" $$\begin{gathered} ∆\mathrm{G}=\sum _{}^{}{\mathrm{n}}_{\mathrm{p}}∆\mathrm{G}{°}_{\mathrm{f}\left(\mathrm{Product}\right)}-∆\mathrm{G}{°}_{\mathrm{f}\left(\mathrm{Reactant}\right)} \\ =\left[\left(1\times 97.89\right)-\left(2\times 51.31\right)\right] \\ =-4.73\mathrm{KJ} \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{\Delta G}°=-{\mathrm{RTlnK}}_{\mathrm{p}} \\ \Rightarrow -4.73=-\left(8.314\times {10}^{-3}\right)\times 298\times {\mathrm{lnK}}_{\mathrm{p}} \\ \Rightarrow -2.47{\mathrm{lnK}}_{\mathrm{p}}=-4.73 \\ \Rightarrow {\mathrm{lnK}}_{\mathrm{p}}=1.91 \\ \Rightarrow {\mathrm{K}}_{\mathrm{p}}={\mathrm{e}}^{1.91} \\ =6.75 \end{gathered}$$

Hence, the equilibrium constant is 6.75.

b)

The pressure for both${\mathrm{NO}}_2$and ${\mathrm{N}}_2{\mathrm{O}}_4$ at $25°\mathrm{C}$is 0.010 atm

$2{\mathrm{NO}}_2\left(\mathrm{g}\right)\rightleftarrows {\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)$

The free energy can be calculated as follows,

$$\begin{gathered} ∆\mathrm{G}=∆\mathrm{G}°+\mathrm{RTlnQ} \\ =-4.73+\left(8.314\times {10}^{-3}\right)\times 298\times \ln \frac{{\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{C}}_4}}{{{\mathrm{P}}^2}_{{\mathrm{NO}}_3}} \\ =-4.73\times 2.478\times \ln \frac{0.01}{{\left(0.01\right)}^2} \\ =6.68\mathrm{KJ}/\mathrm{mol} \end{gathered}$$

So Q=100 is more than ${\mathrm{K}}_{\mathrm{p}}=6.75$, In backward direction the reaction tend to proceed.