A first order reaction has the rate constant, | vedclass

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(D) $A.$ Incorrect. $A$ first-order reaction theoretically takes infinite time to complete.$B.$ Incorrect. $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4

Vedclass · Accepted Answer

$2$

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(D) $A.$ Incorrect. $A$ first-order reaction theoretically takes infinite time to complete.$B.$ Incorrect. $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4.6 	imes 10^{-3}} \approx 150.65 \ s$.$C.$ Incorrect. $t_{10\%} = \frac{1}{k} \ln(\frac{100}{90}) = \frac{2.303}{k} \log(1.11) \approx \frac{0.105}{k}$. $t_{90\%} = \frac{1}{k} \ln(\frac{100}{10}) = \frac{2.303}{k} \log(10) \approx \frac{2.303}{k}$. Thus,$t_{90\%} = 22 	imes t_{10\%}$ (approx),not $25$.$D.$ Correct. For first order,$[A]_t = [A]_0 e^{-kt}$. Degree of dissociation $\alpha = \frac{[A]_0 - [A]_t}{[A]_0} = 1 - e^{-kt}$.$E.$ Correct. For a first-order reaction,rate $= k[A]^1$. Units of rate are $mol \ L^{-1} \ s^{-1}$ and units of $[A]$ are $mol \ L^{-1}$,so units of $k$ are $s^{-1}$. They are not the same.Wait,re-evaluating $E$: Rate unit is $M \ s^{-1}$,$k$ unit is $s^{-1}$. They are different. So $E$ is incorrect.Only statement $D$ is correct. The number of correct statements is $1$.

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