Question

For a chemical reaction, ${\mathrm{K}}_{\mathrm{eq}}\mathrm{is}100\mathrm{at}300\mathrm{K}$ , the value of Δ${\mathrm{G}}^{\mathrm{o}}\mathrm{is}–\mathrm{xR}\mathrm{Joule}\mathrm{at}1\mathrm{atm}\mathrm{pressure}$. Find the value of $\mathrm{x}$. (Use ln $10=2.3$)?

Answer 1

${\mathrm{K}}_{\mathrm{eq}}$ is the equilibrium constant for reversible chemical reaction.

Step 1: Given ${\mathrm{K}}_{\mathrm{eq}}$ $=$ $\mathrm{\Delta G}°=–\mathrm{xR}\mathrm{Joule}$, $R=8.31$${\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}$ , $\ln 10=2.3$

Step 2: The formula used $\mathrm{\Delta G}°=-\mathrm{RTln}\left(\mathrm{Keq}\right)$

Step 3: $\mathrm{\Delta G}°=-\mathrm{RTln}\left(\mathrm{Keq}\right)=-\mathrm{R}\left(300\right)\ln \left(100\right)$

$=$$-\mathrm{R}(300)\left(2\right)\ln \left(10\right)$$=$$-\mathrm{R}(300)\left(2\right)(2.3)$

$=\mathrm{\Delta G}°=-1380\mathrm{R}$

$-\mathrm{xR}=-1380\mathrm{R}$

The value of $\mathrm{x}$ for the chemical reaction is $\mathrm{x}=1380\mathrm{Joule}$