- AHI
- BI2 and P
- CHIO3
- DPI3
Explanation:
Iodination of an alkane is carried out in the presence of oxidizing agent because one of the products of this reaction is hydrogen iodide and this is a strong reducing agent and converts alkyl iodide back to an alkane.
Therefore to counter this difficulty, the iodination of alkanes are carried out in the presence of a strong oxidizing agent like iodic acid (HIO3) which oxidizes the hydrogen iodide formed.
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$A + B \to$ Product
If the concentration of $B$ is increased from $0.1$ to $0.3\, mole$, keeping the value of $A$ at $0.1\, mole$, the rate constant will be

$Pt _{( s )}\left| H _2( g , 1\,atm )\right| H ^{+}( aq , 1 M )|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) \mid Pt ( s )$
When the potential of the cell is $0.712\,V$ at $298\,K$, the ratio $\left[ Fe ^{2+}\right] /\left[ Fe ^{3+}\right]$ is $.......$(Nearest integer)
Given: $Fe ^{3+}+ e ^{-}= Fe ^{2+}, E ^{\circ} Fe ^{3+}, Fe ^{2+} \mid Pt =0.771$
$\frac{2.303 RT }{ F }=0.06\,V$