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  • If the rate constapripof a reaction is $0.03 \mathrm{~s}^{-1}$.

Q.50) If the rate constapripof a reaction is $0.03 \mathrm{~s}^{-1}$. how much time does it take for $7.2 \mathrm{~mol} \mathrm{~L}^{-1}$ concentration of the reactant to get reduced to $0.9 \mathrm{~mol} \mathrm{L-}{ }^{-1}$ ? (Given: $\log 2=0.3$)

A) 21.0 s

B) 69.3 s

C) 23.1 s

D) 210 s

Answers (1)

best_answer

 Given that 

Initial concentration $[R]_0=7.2 \mathrm{~mol} / \mathrm{L}$
Final concentration $[R]=0.9 \mathrm{~mol} / \mathrm{L}$
Rate constant $k=0.03 \mathrm{~s}^{-1}$
$\log 2=0.3$

Since units of $k$ are $\mathrm{s}^{-1}$, this is a first-order reaction.

So, $t=\frac{2.303}{k} \log \left(\frac{[R]_0}{[R]}\right)$

put value 

$\begin{gathered}t=\frac{2.303}{0.03} \log \left(\frac{7.2}{0.9}\right) \\ t=\frac{2.303}{0.03} \log (8)\end{gathered}$

Now,

$$
\begin{aligned}
& \log (8)=\log \left(2^3\right)=3 \log 2=3 \times 0.3=0.9 \\
& t=\frac{2.303}{0.03} \times 0.9=76.77 \times 0.9=\mathbf{6 9 . 0 9 3} \mathrm{s}
\end{aligned}
$$
Hence the correct answer is (2)

Posted by

Saumya Singh

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