A first order reaction has the rate constant, k = 4 . 6 × 10 - 3   s - 1 . The number of correct statement/s from the following is/are Given: log   3 = 0 . 48 . A. Reaction completes in 1000   s . B. The reaction has a half-life of 500   s . C. The time required for 10 %…

Question

A first order reaction has the rate constant, $k = 4.6 \times 10^{- 3} s^{- 1}$ . The number of correct statement/s from the following is/are Given: $\log 3 = 0.48$ .

A. 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.

Aicharya · Accepted Answer

For a first order reaction: $t_{\frac{1}{2}} = \frac{0.693}{k} = \frac{0.693}{4.6 \times 10^{-3} s^{-1}} = 150.65 s$ For a first order reaction, $t_{10\%} = \frac{1}{K} \ln \left(\frac{a}{a-x}\right) = \frac{1}{K} \ln \left(\frac{100}{90}\right)$ $t_{10\%} = \frac{2.303}{K} ( \log 10 - \log 9 )$ $t_{10\%} = \frac{2.093}{K} (0.04)$ Similarly $t_{90\%} = \frac{1}{K} \ln \left(\frac{100}{10}\right)$ $t_{90\%} = \frac{2.303}{K}$ $\frac{t_{90\%}}{t_{10\%}} = \frac{1}{0.04} = 25$

We know,
$e^{kt} = \frac{a}{a - x}$

$\frac{a - x}{a} = e^{- kt}$

$x = a (1 - e^{- kt})$

$α = \frac{x}{a} = (1 - e^{- kt})$

Thus, only option (D) is correct.

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