Question

The gas phase reaction:
$2\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{A}}_2\left(\mathrm{g}\right)$
at 400K has $\mathrm{\Delta G}º=+25.2\mathrm{kJ}{\mathrm{mol}}^{-1}$
The equilibrium constant ${\mathrm{K}}_{\mathrm{c}}$ for this reaction is ______ $\times {10}^{-2}$. (Round off to the Nearest Integer)

[Use $R=8.3\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1},\ln 10=2.3,{\log }_{10}2=0.30,1\mathrm{atm}=1\mathrm{bar}$]
[$antilog\mathit{}(-0.3)=0.501$]

Answer 1

Using the formula,

$$\begin{gathered} \mathrm{DG}°=-{\mathrm{RTlnK}}_{\mathrm{P}} \\ 25200=-2.3\times 8.3\times 400\log \left(K_P\right) \\ {K}_P={10}^{-3.3} \\ ={10}^{-3}\times 0.501 \\ =5.01\times {10}^{-4}{\mathrm{Bar}}^{-1} \\ =5.01\times {10}^{-5}\times {10}^{-4}{\mathrm{Pa}}^{-1} \\ =\frac{K_c}{8.3\times 400} \end{gathered}$$

Finding the value of ${\mathrm{K}}_{\mathrm{c}}$,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{C}}=1.66\times {10}^{-5}{\mathrm{m}}^3/\mathrm{mole} \\ =1.66\times {10}^{-2}\mathrm{L}/\mathrm{mol} \\ =2 \end{gathered}$$

The equilibrium constant ${\mathrm{K}}_{\mathrm{c}}$ for this reaction is $2\times {10}^{-2}$.