Question

A first order reaction is 50 completed in 25 minutes.Calculate the time for 80% completion of the reaction

Answer 1

A first order reaction

t = (1/k) log(A₀/A)

A₀ = initial

A = remaining

for 50% completion A = A₀ - 0.5A₀ = 0.5A₀

25 = (1/k) log(A₀/0.5A₀)

=> 25 = (1/k) log 2

=> 1/k = 25/log2

after 80 % completion A = A₀ - (0.8)A₀ = 0.2A₀

t = (1/k) log(A₀/A)

=> t = (25/Log2) log(A₀/0.2A₀)

=> t = (25/log2)log(10/2)

=> t = 25 ( (log10 - log2)/ log2)

=> t = 25 ( 1 - 0.301)/(0.301)

=> t = 25 (0.699/0.301)

=> t = 58.05

the time for 80% completion of the reaction = 58.05 minuted

Answer 2

Answer:

A first order reaction

t = (1/k) log(A₀/A)

A₀ = initial

A = remaining

for 50% completion A = A₀ - 0.5A₀ = 0.5A₀

25 = (1/k) log(A₀/0.5A₀)

=> 25 = (1/k) log 2

=> 1/k = 25/log2

after 80 % completion A = A₀ - (0.8)A₀ = 0.2A₀

t = (1/k) log(A₀/A)

=> t = (25/Log2) log(A₀/0.2A₀)

=> t = (25/log2)log(10/2)

=> t = 25 ( (log10 - log2)/ log2)

=> t = 25 ( 1 - 0.301)/(0.301)

=> t = 25 (0.699/0.301)

=> t = 58.05

the time for 80% completion of the reaction = 58.05 minuted

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