1
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

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

A
$k=\frac{1}{t} \ln \frac{[A]_t}{[A]_0}$
B
$k=\frac{2303}{t}+\log _{10} \frac{[A]_0}{[A]_t}$
C
$k=-\frac{1}{t} \ln \frac{[A]_t}{[A]_0}$
D
$k=2303 t \log _{10} \frac{[A]_0}{[A]_t}$
2
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-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:

A
$\frac{-3}{2} \frac{d\left[\mathrm{H}_2(\mathrm{~g})\right]}{d t}=\frac{d\left[\mathrm{NH}_3(\mathrm{~g})\right]}{d t}$
B
$\frac{-2}{3} \frac{d\left[\mathrm{H}_2(\mathrm{~g})\right]}{d t}=\frac{d\left[\mathrm{NH}_3(\mathrm{~g})\right]}{\mathrm{dt}}$
C
$\frac{d\left[\mathrm{NH}_3(g)\right]}{d t}=\frac{-1}{3} \frac{d\left[H_2(g)\right]}{d t}$
D
$\frac{-d\left[H_2(g)\right]}{d t}=\frac{d\left[\mathrm{NH}_3(g)\right]}{d t}$
3
MHT CET 2019 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-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

A
$3.2 \times 10^{-6} \mathrm{~s}^{-1}$
B
zero
C
$1.6 \times 10^{-6} \mathrm{~s}^{-1}$
D
$1.6 \times 10^{-5} \mathrm{~s}^{-1}$
4
MHT CET 2019 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

For a chemical reaction rate law is, rate $=k[A]^2[B]$. If $[A]$ is doubled at constant $[B]$, the rate of reaction

A
increases by a factor of 8
B
increases by a factor of 4
C
increases by a factor of 3
D
increases by a factor of 2
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