Find the value of the equilibrium constant (K) of a reaction at 300 K when standard Gibbs free energy change is -25 kH mol -1 ? (Consider R =8.33 J mol -1 K -1 )

Question

Find the value of the equilibrium constant (K) of a reaction at 300 K when standard Gibbs free energy change is -25 kH mol -1 ? (Consider R =8.33 J mol -1 K -1 ) (A) e8 (B) e9 (C) e10 (D) e11

Tardigrade Admin · Accepted Answer

Given, Gibbs free energy =-25 kJ mol -1 Temperature =300 K Equilibrium constant (K)=? Δ G°=-R T ln K -25 kJ mol -1=(-8.314 J mol -1 K -1 × 300 K ) ln k ln K=(-(25 kJ mol -1/8.314 J mol -1 K -1 × 300 K )) =(25 × 103 J mol -1/8.314 J mol -1 K -1 × 300 K ) ln K=-10.02, K=e10.02

Q. Find the value of the equilibrium constant $(K)$ of a reaction at $300 K$ when standard
Gibbs free energy change is $- 25 k H m o l^{- 1} ?$
(Consider $R = 8.33 J m o l^{- 1} K^{- 1}$ )

$e^{8}$

$e^{9}$

$e^{10}$

$e^{11}$

Q. Find the value of the equilibrium constant (K) of a reaction at 300K when standard Gibbs free energy change is −25kHmol−1? (Consider R=8.33Jmol−1K−1 )

e8

e9

e10

e11

Given, Gibbs free energy =−25kJmol−1 Temperature =300K Equilibrium constant (K)=? ΔG∘=−RTlnK −25kJmol−1=(−8.314Jmol−1K−1×300K)lnk lnK=(−8.314Jmol−1K−1×300K25kJmol−1​) =8.314Jmol−1K−1×300K25×103Jmol−1​ lnK=−10.02,K=e10.02

Q. Find the value of the equilibrium constant $(K)$ of a reaction at $300 K$ when standard
Gibbs free energy change is $- 25 k H m o l^{- 1} ?$
(Consider $R = 8.33 J m o l^{- 1} K^{- 1}$ )

$e^{8}$

$e^{9}$

$e^{10}$

$e^{11}$