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Question: A first order reaction : B===>C has a half life of 20mins. What percentage shall have reacted after 47minutes

Sagot :

Answer:19.6%

Explanation:

From K=0.693/t

0.693/20 =0.03465min^-1

But rate law is K =2.303/t log a/(a-x)

Substitute the value of k

0.03465=2.303/47 log a / (a-x)

Log a/ (a-x) = 0.7071

a/ (a-x) = 5.094

(a-x)/ a = 1/5.094 = 0.196

Percentage of reactants remaining after 47 minutes= 0.196× 100% = 19.6%

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