Which of the following relations is correct f | vedclass

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(B) For a first order reaction,the time $t_x$ required to complete a fraction $x$ is given by $t_x = \frac{1}{k} \ln(\frac{1}{1-x})$.For $t_{1/2}$,$x

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$t_{7/8} / t_{1/2} = 3$ (first order reaction)

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(B) For a first order reaction,the time $t_x$ required to complete a fraction $x$ is given by $t_x = \frac{1}{k} \ln(\frac{1}{1-x})$.For $t_{1/2}$,$x = 1/2$,so $t_{1/2} = \frac{\ln 2}{k}$.For $t_{3/4}$,$x = 3/4$,so $t_{3/4} = \frac{1}{k} \ln(\frac{1}{1-3/4}) = \frac{\ln 4}{k} = \frac{2 \ln 2}{k} = 2 t_{1/2}$. Thus,$t_{3/4} / t_{1/2} = 2$.For $t_{7/8}$,$x = 7/8$,so $t_{7/8} = \frac{1}{k} \ln(\frac{1}{1-7/8}) = \frac{\ln 8}{k} = \frac{3 \ln 2}{k} = 3 t_{1/2}$. Thus,$t_{7/8} / t_{1/2} = 3$.For $t_{15/16}$,$x = 15/16$,so $t_{15/16} = \frac{1}{k} \ln(\frac{1}{1-15/16}) = \frac{\ln 16}{k} = \frac{4 \ln 2}{k} = 4 t_{1/2}$. Thus,$t_{15/16} / t_{1/2} = 4$.All options $A, B, D$ are correct for first order reactions. However,in standard multiple-choice formats,$B$ is a commonly cited identity.

Which of the following relations is correct for the given reaction orders?

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