Calculate the amount of reactant in percent t | vedclass

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(A) For a first-order reaction,the rate constant is given by: $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right)$ Given $k = 0.02303 \ min^{-1

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) For a first-order reaction,the rate constant is given by: $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} ight)$ Given $k = 0.02303 \ min^{-1}$ and $t = 60 \ min$. Assuming initial concentration $[A]_0 = 100$,we need to find $[A]$. $0.02303 = \frac{2.303}{60} \log \left( \frac{100}{[A]} ight)$ $0.02303 	imes \frac{60}{2.303} = \log \left( \frac{100}{[A]} ight)$ $0.01 	imes 60 = \log \left( \frac{100}{[A]} ight)$ $0.6 = \log \left( \frac{100}{[A]} ight)$ Since $\log(4) \approx 0.602$,we have $\frac{100}{[A]} = 4$,so $[A] = 25$. Alternatively,using half-life: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02303} \approx 30 \ min$. After $60 \ min$ $(2 	imes t_{1/2})$,the amount remaining is $(\frac{1}{2})^2 	imes 100 \% = 25 \%$.

Calculate the amount of reactant in percent that remains after $60 \ min$ in a first-order reaction. $(k = 0.02303 \ min^{-1})$ (in $\%$)

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