For the first order reaction A rightarrow B,t | vedclass

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(B) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k} = 30 \ min$.The time required for $75 \%$ completion $(T_{75\%})$

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$60$

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(B) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k} = 30 \ min$.The time required for $75 \%$ completion $(T_{75\%})$ is the time taken for two half-lives to pass.$T_{75\%} = 2 	imes t_{1/2} = 2 	imes 30 \ min = 60 \ min$.Alternatively,using the formula $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$:$t = \frac{2.303}{k} \log \frac{100}{100-75} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} 	imes 2 \log 2$.Since $t_{1/2} = \frac{2.303 \log 2}{k} = 30 \ min$,then $t = 2 	imes 30 = 60 \ min$.

For the first order reaction $A \rightarrow B$,the half-life is $30 \ min$. The time taken for $75 \%$ completion of the reaction is $..... \ min$. (Nearest integer)
Given: $\log 2 = 0.3010, \log 3 = 0.4771, \log 5 = 0.6989$

Similar Questions

Consider the following two first order reactions occurring at $298 \ K$ with same initial concentration of $A$: $(1)$ $A \rightarrow B$; rate constant, $k=0.693 \ min^{-1}$ $(2)$ $A \rightarrow C$; half-life, $t_{1/2}=0.693 \ min$. Choose the correct option.

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If the half-life of a first-order reaction is $6.93$ minutes, then the time required for the completion of $99\%$ of the reaction will be . . . . . . minutes. (Given: $\log 2 = 0.3010$)

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For the first order reaction $A \rightarrow B$,the half-life is $30 \ min$. The time taken for $75 \%$ completion of the reaction is $..... \ min$. (Nearest integer)Given: $\log 2 = 0.3010, \log 3 = 0.4771, \log 5 = 0.6989$