The half-life of a certain first-order reaction is 15 minutes. What fraction of the original reactant concentration will remain after 2.0 hours?

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Answer 1 · 2018-05-26T21:58:35

First-order half-life does not depend on the initial concentration

#t_(1//2) = (ln2)/k#,

and the rate constant does not depend on the current concentration, so the half-life will not change over time.

Therefore, #n# number of half-lives passing by will decrease the concentration to #(1/2)^n# as much.

As a result, if #t_(1//2) = "15 min"#, and #"2.0 hours"# passed...

#"2.0 hours"/"15 min" = "8 half-lives"#

And thus, #(1/2)^8 = color(blue)(1/256)# of the starting concentration is left.