TG EAPCET 2024 (Online) 11th May Morning Shift | Chemical Kinetics Question 1 | Chemistry

$ \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 $K_{3}$

$K_{1}=2 K_{2}=4 K_{3}$

$2 K_{1}=K_{2}=4 K_{3}$

$2 K_{1}=3 K_{2}=4 K_{3}$

$4 K_{1}=2 K_{2}=K_{3}$

TG EAPCET 2024 (Online) 10th May Evening Shift

MCQ (Single Correct Answer)

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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 $\mathrm{min}^{-1}$ )?

$2 \times 10^{-3}$

$\frac{2 \times 10^{-3}}{2.303}$

$4.606 \times 10^{-3}$

$0.5 \times 10^{-5}$

TG EAPCET 2024 (Online) 10th May Morning Shift

MCQ (Single Correct Answer)

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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: $\log 2=0.3$ )

TG EAPCET 2024 (Online) 9th May Evening Shift

MCQ (Single Correct Answer)

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Consider the gaseous reaction, $$ A_2+B_2 \longrightarrow 2 A B $$ The following data was obtained for the above reaction.

[A₂]₀	[B₂]₀	Initial rate of formation of AB (mol L⁻¹ s⁻¹)
0.1 M	0.1 M	2.5 × 10⁻⁴
0.2 M	0.1 M	5.0 × 10⁻⁴
0.2 M	0.1 M	1.0 × 10⁻³

The value of rate constant for the above reaction is

$1.25 \times 10^{-2}$

$1.25 \times 10^{-3}$

$2.5 \times 10^{-2}$

$2.5 \times 10^{-1}$

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