Answer:
![\Delta{G^0}_2 =-294.25\ kJ/mol](https://tex.z-dn.net/?f=%5CDelta%7BG%5E0%7D_2%20%3D-294.25%5C%20kJ%2Fmol)
Explanation:
The relation between standard Gibbs energy and equilibrium reaction is shown below as:
![\Delta{G^0} =-RT \ln K](https://tex.z-dn.net/?f=%5CDelta%7BG%5E0%7D%20%3D-RT%20%5Cln%20K)
R is Gas constant having value = 0.008314 kJ / K mol
Given temperature, T = 300 K (Source Original)
Given, ![\Delta{G^0}_1=-300\ kJ/mol](https://tex.z-dn.net/?f=%5CDelta%7BG%5E0%7D_1%3D-300%5C%20kJ%2Fmol)
So,
![-300=-0.008314\times 300 \ln K](https://tex.z-dn.net/?f=-300%3D-0.008314%5Ctimes%20300%20%5Cln%20K)
![\ln \left(K\right)=\frac{300}{2.4942}](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28K%5Cright%29%3D%5Cfrac%7B300%7D%7B2.4942%7D)
![\ln \left(K\right)=120.27904](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28K%5Cright%29%3D120.27904)
Also,
K₁ = 10*K₂
K₂ = 0.1 K₁
So,
![\Delta{G^0}_2 =-0.008314\times 300 \ln (0.1\times K_1)](https://tex.z-dn.net/?f=%5CDelta%7BG%5E0%7D_2%20%3D-0.008314%5Ctimes%20300%20%5Cln%20%280.1%5Ctimes%20K_1%29)
![\Delta{G^0}_2 =-0.008314\times 300 (\ln 0.1+ \ln K_1)](https://tex.z-dn.net/?f=%5CDelta%7BG%5E0%7D_2%20%3D-0.008314%5Ctimes%20300%20%28%5Cln%200.1%2B%20%5Cln%20K_1%29)
![\Delta{G^0}_2 =-0.008314\times 300 (-2.3026+120.27904)](https://tex.z-dn.net/?f=%5CDelta%7BG%5E0%7D_2%20%3D-0.008314%5Ctimes%20300%20%28-2.3026%2B120.27904%29)
![\Delta{G^0}_2 =-294.25\ kJ/mol](https://tex.z-dn.net/?f=%5CDelta%7BG%5E0%7D_2%20%3D-294.25%5C%20kJ%2Fmol)
A because KCL 2 and K + CL cancel out and Both of the Ca cancel each other out
Too kill everyone but sometimes they don’t even think about that so