1
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Observe the following data given in the table ( $K_H=$ Henry's law constant)

$$ \begin{aligned} &\begin{array}{ccccc} \hline \text { Gas } & \mathrm{CO}_2 & \mathrm{Ar} & \mathrm{HCHO} & \mathrm{CH}_4 \\ \hline\left(\boldsymbol{K}_{\mathrm{H}} \text { bar at } \mathbf{2 9 8 ~ K}\right) & 1.67 & 40.3 & 1.83 \times 10^{-5} & 0.413 \\ \hline \end{array}\\ &\text { The correct order of their solubility in water is } \end{aligned} $$

A

$\mathrm{CO}_2>\mathrm{CH}_4>\mathrm{HCHO}>\mathrm{Ar}$

B

$\mathrm{Ar}>\mathrm{HCHO}>\mathrm{CH}_4>\mathrm{CO}_2$

C

$\mathrm{HCHO}>\mathrm{CH}_4>\mathrm{CO}_2>\mathrm{Ar}$

D

$\mathrm{CO}_2>\mathrm{HCHO}>\mathrm{CH}_4>\mathrm{Ar}$

2
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The Gibbs energy change of the reaction (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) corresponding to the following cell

$\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0.1 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.001 \mathrm{M})\right| \mathrm{Fe}$

(Given $E_{\mathrm{Cr}^{3+} \mid \mathrm{Cr}}^{\circ}=-0.75 \mathrm{~V} ; E_{\mathrm{Fe}^{2+} \mid \mathrm{Fe}}^{\circ}=-0.45 \mathrm{~V}$,

$\left.\mathrm{IF}=96,500 \mathrm{C} \mathrm{mol}^{-1}\right)$

A

-150.9

B

-173.7

C

+150.9

D

+173.7

3
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

For a first order decomposition of a certain reaction, rate constant is given by the equation. $\log k\left(s^{-1}\right)=7.14-\frac{1 \times 10^4 \mathrm{~K}}{T}$. The activation energy of the reaction ( in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is

$$ \left(R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) $$

A

161.1

B

171.1

C

181.1

D

191.1

4
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The source of an enzyme is malt and that enzyme converts $X$ into $Y . X$ and $Y$ respectively are

A

starch, maltose

B

maltose, glucose

C

proteins, peptides

D

glucose, fructose

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