1
TS EAMCET 2023 (Online) 14th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A possible mechanism for the gaseous reaction $2 \mathrm{H}_2+2 \mathrm{NO} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{N}_2$ is

Step 1:2 $\mathrm{NO} \rightleftharpoons \mathrm{N}_2 \mathrm{O}_2$

Step 2 : $\mathrm{N}_2 \mathrm{O}_2+\mathrm{H}_2 \longrightarrow \mathrm{~N}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O}$ (slow)

Step 3: $\mathrm{N}_2 \mathrm{O}+\mathrm{H}_2 \longrightarrow \mathrm{~N}_2+\mathrm{H}_2 \mathrm{O}$

The rate law for this reaction is

A

$R=k[\mathrm{NO}]^2\left[\mathrm{H}_2\right]^2$

B

$R=k[\mathrm{NO}]\left[\mathrm{H}_2\right]^2$

C

$R=k[\mathrm{NO}]^{1 / 2}\left[\mathrm{H}_2\right]$

D

$R=k[\mathrm{NO}]^2\left[\mathrm{H}_2\right]$

2
TS EAMCET 2023 (Online) 14th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The reduction potential of a half-cell consisting of a Pt electrode immersed in $2.0 \mathrm{M} \mathrm{Fe}^{2+}$ and $0.02 \mathrm{M} \mathrm{Fe}^{3+}$ solution (in V) is

Given : $\left(\frac{2.303 R T}{F}=0.059, E_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^{\circ}=0.771 \mathrm{~V}\right)$

A

0.543

B

0.653

C

0.733

D

0.822

3
TS EAMCET 2023 (Online) 14th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The sol prepared by Bredig's arc method is $X$ and the charge of sol particles of it is $q . X$ and $q$ are respectively

A

metal sol, -ve

B

metal sol, $+v e$

C

metal sulphide sol, -ve

D

$\mathrm{TiO}_2$ sol, +ve

4
TS EAMCET 2023 (Online) 14th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The metal which is refined by Mond process is $(X)$, by van Arkel process is $(Y)$ and by zone refining is $(Z), X, Y$ and Z respectively are

A

$\mathrm{Ni}, \mathrm{Zr}, \mathrm{Ga}$

B

$\mathrm{Zr}, \mathrm{Ni}, \mathrm{Ga}$

C

$\mathrm{Ga}, \mathrm{Ni}, \mathrm{Zr}$

D

$\mathrm{Ni}, \mathrm{Ga}, \mathrm{Zr}$

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