11. ⇒ (MHT CET 2021 21th September Evening Shift )

For a first order reaction, intercept of the graph between $\log \frac{[A]_{o}}{[A]_{t}}$ (Y-axis) and conc. (X-axis) is equal to

A. $- \frac{k}{2.303 K}$

B. $- \log [A]_{o}$

C. zero

D. $\frac{2.303}{K}$

Correct Option is (C)

Currently no explanation available

12. ⇒ (MHT CET 2021 21th September Evening Shift )

What is the half-life of a first order reaction if time required to decrease concentration of reactant from 1.0 M to 0.25 M is 10 hour?

A. 12 hour

B. 4 hour

C. 5 hour

D. 10 hour

Correct Option is (C)

$t = 10$ hour, $[A]_{0} = 1.0 M, [A]_{t} = 0.25 M$

For first order reaction,

(i) $\begin{aligned} k & = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}} & = \frac{2.303}{10 h} \log_{10} \frac{1.0}{0.25} \\ = \frac{2.303}{10 h} \log_{10} 4 & = \frac{2.303 \times 0.6020}{10 h} \\ = 0.139 h^{- 1} \end{aligned}$

(ii) $\begin{aligned} t_{1 / 2} & = \frac{0.693}{k} & = \frac{0.693}{0.139 h^{- 1}} \\ = 4.98 hour & \approx 5 hours \end{aligned}$

13. ⇒ (MHT CET 2021 21th September Morning Shift )

In a first order reaction, concentration of reactant is reduced to (1/8)^th of concentration in 23.03 minutes. What is half-life period of reaction?

A. 25 min

B. 7.7 min

C. 15 min

D. 30 min

Correct Option is (B)

$[A]_{0} = 1, [A]_{t} = \frac{1}{8}, t = 23.03 \min$

For first order reaction,

$\begin{aligned} (i) k & = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}} \\ = \frac{2.303}{23.03 min} \log_{10} \frac{1}{(1 / 8)} \\ = \frac{2.303}{23.03} \times \log_{10} 8 = \frac{2.303 \times 0.9031}{23.03} \\ = 0.0903 \min^{- 1} \end{aligned}$

(ii) $t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0903 \min^{- 1}} = 7.7 \min$

14. ⇒ (MHT CET 2021 20th September Evening Shift )

In a first order reaction concentration of reactant decreases from 20 m mol to 10 m mol in 1.151 min. What is rate constant?

A. 1.15 min $^{- 1}$

B. 3.0 min $^{- 1}$

C. 5.50 min $^{- 1}$

D. 0.60 min $^{- 1}$

Correct Option is (D)

$[A]_{0} = 20 m mol, [A]_{t} = 10 m mol$

$t = 1.151 \min, k =$ ?

For first order reaction,

$k = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}}$

$∴ k = \frac{2.303}{1.151 \min^{- 1}} \log_{10} (\frac{20}{10}) = \frac{2.303 \times \log 2}{1.151} = \frac{2.303 \times 0.3010}{1.151} = 0.60 \min^{- 1}$

15. ⇒ (MHT CET 2021 20th September Morning Shift )

Half life for a first order reaction is 6.93 hour. What is the time required for 80% completion of the reaction?

A. 12 hours

B. 18 hours

C. 6 hours

D. 16 hours

Correct Option is (D)

For first order reaction,

$k = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{6.93} = 0.1 {hour}^{- 1}$

Here, $[A]_{0} = 100, [A]_{t} = 100 - 80 = 20$

$\begin{aligned} Now, k = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}} \\ ∴ t = \frac{2.303}{0.1} \log_{10} \frac{100}{20} = 23.03 \times \log_{10} 5 = 23.03 \times 0.699 = 16.10 hours. \end{aligned}$

⇒Chemical Kinetics