SATHEE: Chemical Kinetics - Result Question 21

Chemical Kinetics - Result Question 21

####21. A first order reaction has a specific reaction rate of $10^{- 2} s^{- 1}$ . How much time will it take for $20 g$ of the reactant to reduce to $5 g$ ?

(a) $138.6 s e c$

(b) $346.5 s e c$

(d) $238.6 s e c$

[2017]

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Solution:

(a) Half life for a first order reaction,

$\begin{aligned} t_{1 / 2} = \frac{0.693}{K} \\ So, t_{1 / 2} = \frac{0.693}{10^{- 2}} s e c \end{aligned}$

Also, for the reduction of $20 g$ of reactant to $5 g$ , two half lives will be required.

$∴$ For $20 g$ of the reactant to reduce to $5 g$ , timetaken,

$t = 2 \times \frac{0.693}{10^{- 2}} s e c = 138.6 s e c$ .

$N = N_{0} (\frac{1}{2})^{n}$

where $N =$ mass after $n$ number of half life

$n =$ number of half life

$N_{0} =$ Initital mass

Chemical Kinetics - Result Question 21

Solution:

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Chemical Kinetics - Result Question 21

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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