Question

For the reaction $\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{B}\left(\mathrm{g}\right)$ at $495\mathrm{K}$ ;:${\mathrm{\Delta rG}}_{\mathrm{o}}=–9.478\mathrm{kJ}{\mathrm{mol}}^{-1}$

If we start the reaction in a closed container at $495\mathrm{K}$ with $22$millimoles of $\mathrm{A}$, the amount of $\mathrm{B}$ in the equilibrium mixture is ________millimoles. (Round off to the nearest Integer).

[$\mathrm{R}=8.314{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1};\ln 10=2.303$]

Answer 1

The above question can be solved as:

Step 1:

$$\begin{gathered} \mathrm{\Delta G}°=–\mathrm{RT}\ln \mathrm{Keq} \\ =–9.478\times 103 \\ =–495\times 8.314\ln \mathrm{Keq} \\ \ln \mathrm{Keq}=2.303 \\ =\ln 10 \\ \mathrm{So},\mathrm{Keq}=10 \end{gathered}$$

Step 2: Now,

$$\begin{gathered} \mathrm{A}\left(\mathrm{g}\right)\underset{}{\overset{}{\to }}\mathrm{B}\left(\mathrm{g}\right) \\ \mathrm{t}=0220 \\ \mathrm{t}=\mathrm{t}22–\mathrm{x}\mathrm{x} \end{gathered}$$

Step 3:

$$\begin{gathered} \mathrm{Keq}=\left[\mathrm{B}\right]/\left[\mathrm{A}\right] \\ 10=\mathrm{X}/(22-\mathrm{X}) \\ 10\mathrm{x}=20 \end{gathered}$$

So, millimoles of $\mathrm{B}=20$

Therefore, the amount of $\mathrm{B}$ in the equilibrium mixture is $20$ millimoles.