Answer:
[tex]\Delta{G^0}_2 =-294.25\ kJ/mol[/tex]
Explanation:
The relation between standard Gibbs energy and equilibrium reaction is shown below as:
[tex]\Delta{G^0} =-RT \ln K[/tex]
R is Gas constant having value = 0.008314 kJ / K mol
Given temperature, T = 300 K (Source Original)
Given, [tex]\Delta{G^0}_1=-300\ kJ/mol[/tex]
So,
[tex]-300=-0.008314\times 300 \ln K[/tex]
[tex]\ln \left(K\right)=\frac{300}{2.4942}[/tex]
[tex]\ln \left(K\right)=120.27904[/tex]
Also,
K₁ = 10*K₂
K₂ = 0.1 K₁
So,
[tex]\Delta{G^0}_2 =-0.008314\times 300 \ln (0.1\times K_1)[/tex]
[tex]\Delta{G^0}_2 =-0.008314\times 300 (\ln 0.1+ \ln K_1)[/tex]
[tex]\Delta{G^0}_2 =-0.008314\times 300 (-2.3026+120.27904)[/tex]
[tex]\Delta{G^0}_2 =-294.25\ kJ/mol[/tex]
