MHT CET 2020 16th October Morning Shift | Chemical Kinetics Question 74 | Chemistry

A first order reaction has rate constant $$1 \times 10^{-2} \mathrm{~s}^{-1}$$. What time will, it take for $$20 \mathrm{~g}$$ or reactant to reduce to $$5 \mathrm{~g}$$ ?

693.0 s

138.6 s

238.6 s

346.5 s

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

The integrated rate equation for first order reaction, $A \rightarrow$ product, is

$k=\frac{1}{t} \ln \frac{[A]_t}{[A]_0}$

$k=\frac{2303}{t}+\log _{10} \frac{[A]_0}{[A]_t}$

$k=-\frac{1}{t} \ln \frac{[A]_t}{[A]_0}$

$k=2303 t \log _{10} \frac{[A]_0}{[A]_t}$

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

For the elementary reaction, $3 \mathrm{H}_2(g)+\mathrm{N}_2(g) \longrightarrow 2 \mathrm{NH}_3(g)$ identify the correct relation among the following relations:

$\frac{-3}{2} \frac{d\left[\mathrm{H}_2(\mathrm{~g})\right]}{d t}=\frac{d\left[\mathrm{NH}_3(\mathrm{~g})\right]}{d t}$

$\frac{-2}{3} \frac{d\left[\mathrm{H}_2(\mathrm{~g})\right]}{d t}=\frac{d\left[\mathrm{NH}_3(\mathrm{~g})\right]}{\mathrm{dt}}$

$\frac{d\left[\mathrm{NH}_3(g)\right]}{d t}=\frac{-1}{3} \frac{d\left[H_2(g)\right]}{d t}$

$\frac{-d\left[H_2(g)\right]}{d t}=\frac{d\left[\mathrm{NH}_3(g)\right]}{d t}$

MHT CET 2019 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

The activation energy of a reaction is zero. Its rate constant at 280 K is $1.6 \times 10^{-6} \mathrm{~s}^{-1}$, the rate constant at 300 K is

$3.2 \times 10^{-6} \mathrm{~s}^{-1}$

zero

$1.6 \times 10^{-6} \mathrm{~s}^{-1}$

$1.6 \times 10^{-5} \mathrm{~s}^{-1}$

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