SATHEE: Chemical Kinetics - Result Question 18

Chemical Kinetics - Result Question 18

####18. A first order reaction has a rate constant of $2.303 \times 10^{- 3} s^{- 1}$ . The time required for $40 g$ of this reactant to reduce to $10 g$ will be

[Given that $\log_{10} 2 = 0.3010$ ]

[NEET Odisha 2019]

(a) $602 s$

(b) $230.3 s$

(d) $2000 s$

Show Answer

Solution:

(a) For a first order reaction

Half life period, $t_{\frac{1}{2}} = \frac{0.693}{k} = \frac{0.693}{2.303 \times 10^{- 3}} s^{- 1}$ $= 300.91 s$

Now, $40 g \overset{t_{1 / 2}}{\to} 20 g \overset{t_{1 / 2}}{\to} 10 g$

So, $40 g$ substance requires 2 half life periods to reduce upto $10 g$

$∴$ Time taken in reduction $= 2 \times 300.91 s$

$= 601.82 ≃ 602 s$

Chemical Kinetics - Result Question 18

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Chemical Kinetics - Result Question 18

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu