6. ⇒ (MHT CET 2023 10th May Evening Shift )

Time required for $90 %$ completion of a first order reaction is ' $x$ ' minute. Calculate the time required to complete $99.9 %$ of the reaction at same temperature.

A. $x$ minute

B. $2 x$ minute

C. $3 x$ minute

D. $\frac{x}{2}$ minute

Correct Option is (C)

$\begin{aligned} t_{90 % 6} & = \frac{2.303}{k} \log_{10} \frac{[A]_{0}}{[A]_{t}} = \frac{2.303}{k} \log_{10} \frac{100}{10} \\ = \frac{2.303}{k} \log_{10} 10 \\ t_{99.9 %} & = \frac{2.303}{k} \log_{10} \frac{[A]_{0}}{[A]_{t}} = \frac{2.303}{k} \log_{10} \frac{100}{0.1} \\ = \frac{2.303}{k} \log_{10} 1000 \\ \frac{t_{999 %}}{t_{90 %}} & = \frac{\frac{2.303}{k} \log_{10} 1000}{\frac{2.303}{k} \log_{10} 10} = \frac{\log_{10} 1000}{\log_{10} 10} = \frac{3}{1} \\ ∴ t_{99.9 %} & = 3 \times t_{90 %} = 3 x minute (since, t_{90 %} = x minute) \end{aligned}$

7. ⇒ (MHT CET 2023 10th May Morning Shift )

What is the half life of a first order reaction if rate constant is $4.2 \times 10^{- 2}$ per day?

A. 5.0 day

B. 16.5 day

C. 28.0 day

D. 9.0 day

Correct Option is (B)

For a first order reaction,

$t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{4.2 \times 10^{- 2}} = 16.5$ days

8. ⇒ (MHT CET 2023 9th May Evening Shift )

Calculate the rate constant of first order reaction if the concentration of the reactant decreases by $90 %$ in 30 minutes.

A. $7.7 \times 10^{- 2}$ minute $^{- 1}$

B. $4.2 \times 10^{- 2}$ minute $^{- 1}$

C. $2.1 \times 10^{- 2}$ minute $^{- 1}$

D. $3.5 \times 10^{- 2}$ minute $^{- 1}$

Correct Option is (A)

Concentration decreases by $90 %$ . Hence, $10 %$ reactant is left after 30 minutes.

$\begin{aligned} k & = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}} \\ k & = \frac{2.303}{30} \log_{10} \frac{100}{10} \\ = \frac{2.303}{30 \min} \\ = 7.7 \times 10^{- 2} \min^{- 1} \end{aligned}$

9. ⇒ (MHT CET 2023 9th May Evening Shift )

The rate law for the reaction $A + B \to$ product is given by rate $= k [A] [B]$ Calculate $[A]$ if rate of reaction and rate constant are $0.25 mol {dm}^{- 3} s^{- 1}$ and $6.25 {mol}^{- 1} {dm}^{3} s^{- 1}$ respectively $[[B] = 0.25 mol {dm}^{- 3}]$

A. $0.22 mol {dm}^{3}$

B. $0.16 mol {dm}^{3}$

C. $0.30 mol {dm}^{3}$

D. $0.25 mol {dm}^{3}$

Correct Option is (B)

$\begin{aligned} \begin{aligned} rate = k [A] [B] \\ ∴ [A] = \frac{rate}{k [B]} & = \frac{0.25 mol {dm}^{- 3} s^{- 1}}{6.25 {mol}^{- 1} {dm}^{3} \cdot s^{- 1} \times 0.25 mol {dm}^{- 3}} \\ = 0.16 mol {dm}^{3} \end{aligned} \end{aligned}$

10. ⇒ (MHT CET 2023 9th May Morning Shift )

Calculate the amount of reactant in percent that remains after 60 minutes involved in first order reaction. $(k = 0.02303$ minute $^{- 1})$

A. $25 %$

B. $75 %$

C. $50 %$

D. $12.5 %$

Correct Option is (A)

For a first order reaction,

$\begin{aligned} k = \frac{0.693}{t_{1 / 2}} \\ t_{1 / 2} = \frac{0.693}{0.02303} = 30 \min \end{aligned}$

Percent of reactant that remains after $t_{1 / 2} = 50 %$

$2 \times t_{1 / 2} = 60 \min$

Therefore, percent of reactant that remains after $2 t_{1 / 2} = 25 %$

⇒Chemical Kinetics