A first order reaction has the rate constant, k=4.6×10−3s−1 . The number of correct statement/s from thefollowing is/are __________Given : log 3 = 0.48A. Reaction completes in 1000 s.B. The reaction has a half-life of 500 s.C. The time required for 10% completion is 25 times the time required for 90% completion.D. The degree of dissociation is equal to (1-e-kt)E. The rate and the rate constant have the same unit

Solution:

$A . t = \frac{2.303}{K} \log \frac{a}{(a - x)} \Rightarrow t = \frac{2.303}{4.6 \times 10^{- 3}} = \frac{2.303}{4.6} \times 10^{3} = 2500 \sec B . t_{1 / 2} = \frac{0.693}{K} = \frac{0.693}{4.6} \times 10^{3} = \frac{693}{4.6} = 150 \sec \frac{C . t_{90 %}}{t_{10 %}} = \frac{\frac{2.303}{K} \log (\frac{100}{10})}{\frac{2.303}{k} \log (\frac{100}{10})} = \frac{\log 10}{\log 10 - \log 9} = \frac{1}{0.04} = 25$
D. Arrhenious equation $K = A_{0} e^{- kt}$
$\begin{array}{l} A \to B \\ a_{0} 0 \\ a_{0} (1 - α) a_{a} α \\ a_{0} (1 - α) = a \cdot e^{- kt} \\ (1 - α) = e^{- kt} \\ α = e^{1 - kt} \\ = 1 - e^{- kt} \end{array}$

Solution:

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