1
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

For the reaction $R \rightarrow P$, half life is independent of initial concentration of the reactant, $R$. Which one of the following graphs is not correct for the reaction?

A
TG EAPCET 2025 (Online) 2nd May Morning Shift Chemistry - Chemical Kinetics Question 22 English Option 1
B
TG EAPCET 2025 (Online) 2nd May Morning Shift Chemistry - Chemical Kinetics Question 22 English Option 2
C
TG EAPCET 2025 (Online) 2nd May Morning Shift Chemistry - Chemical Kinetics Question 22 English Option 3
D
TG EAPCET 2025 (Online) 2nd May Morning Shift Chemistry - Chemical Kinetics Question 22 English Option 4
2
TG EAPCET 2024 (Online) 11th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

For the gaseous reaction, $\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$

the rate can be expressed as

$ \begin{array}{l} -\frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=K_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \\\\ +\frac{d\left[\mathrm{NO}_{2}\right]}{d t}=K_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \\\\ +\frac{d\left[\mathrm{O}_{2}\right]}{d t}=K_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \end{array} $

The correct relation between $K_{1}, K_{2}$ and $K_{3}$

A
$K_{1}=2 K_{2}=4 K_{3}$
B
$2 K_{1}=K_{2}=4 K_{3}$
C
$2 K_{1}=3 K_{2}=4 K_{3}$
D
$4 K_{1}=2 K_{2}=K_{3}$
3
TG EAPCET 2024 (Online) 10th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
For a first order reaction, the graph between $\log \frac{a}{(a-x)}$ (on $y$-axis) and time (in min, on $x$-axis) gave a straight line passing through origin. The slope is $2 \times 10^{-3} \mathrm{~min}^{-1}$. What is the rate constant (in $\mathrm{min}^{-1}$ )?
A
$2 \times 10^{-3}$
B
$\frac{2 \times 10^{-3}}{2.303}$
C
$4.606 \times 10^{-3}$
D
$0.5 \times 10^{-5}$
4
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The decomposition of benzene diazonium chloride is a first order reaction. The time taken for its decomposition to $\frac{1}{4}$ and $\frac{1}{10}$ of its initial concentration are $t_{\frac{1}{4}}$ and $t_{\frac{1}{4}}^{10}$ respectively. The value of $\frac{t_{\frac{1}{4}}^4}{t_1} \times 100$ is (Give: $\log 2=0.3$ )
A
60
B
30
C
90
D
45

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