Answer:
[tex]K_T=2.014x10^6[/tex]
Explanation:
Hello,
By assuming the validity (constant ΔH°rxn) of the van't Hoff equation:
[tex]ln(\frac{K_T}{K_{T^0}})=-\frac{H_{rxn}}{R} (\frac{1}{T} -\frac{1}{T^0} )[/tex]
Solving for [tex]K_T[/tex] it means the equilibrium constant at 1 °C, we get:
[tex]ln(K_T)=ln(2.25x10^4)-\frac{-128000J/mol}{8.314J/(mol*K)}*(\frac{1}{274.15K}-\frac{1}{298K} ) \\\\ln(K_T)=14.52[/tex]
Finally:
[tex]K_T=exp(14.52)\\\\K_T=2.014x10^6[/tex]
Best regards.