Answer:
[tex]k=1.12x10^{-4}s[/tex]
Explanation:
Hello,
By considering the rate law:
[tex]ln(\frac{[CO_2]}{[CO_2]_0} )=-kt\\[/tex]
Solving for the time, we've got:
[tex]k=-\frac{ln (\frac{[CO_2]}{[CO_2]_0 })}{t}\\\\k=-\frac{ln(\frac{56.0mmol*dm^{-3} }{220mmol*dm^{-3} })}{1.22x10^{4}s }\\ \\k=1.12x10^{-4}s[/tex]
Best regards.