The rusting of iron takes place as follows RCOCl+H2 xrightarrowPd/BaSO4RCHO+HCl BaSO4 Calculate HCl for the net process.

Question

The rusting of iron takes place as follows RCOCl+H2 xrightarrowPd/BaSO4RCHO+HCl BaSO4 Calculate HCl for the net process. (A)  l>II>lll>IV  (B)  IV>ll>lll>l  (C)  lll>IV>H>l  (D)  lll>ll>IV>l

Tardigrade Admin · Accepted Answer

Reactions, (i) Cl- and 825× 1014M (ii) 1.65× 1013M and 2.87× 106M Net reaction 6.72× 106M Kf -0.360oC =-0.88-2.46 =-3.34F=-3.34× 96500J =-322.31kJ=-322kJ

Q. The rusting of iron takes place as follows $RCOCl + H_{2} P d / B a S O_{4} RC H O + H Cl$ $B a S O_{4}$ Calculate $H Cl$ for the net process.

$l > II > lll > I V$

$I V > ll > lll > l$

$lll > I V > H > l$

$lll > ll > I V > l$

Reactions, (i) $C l^{-}$ and $825 \times 10^{14} M$ (ii) $1.65 \times 10^{13} M$ and $2.87 \times 10^{6} M$ Net reaction $6.72 \times 10^{6} M$ $K_{f}$ $- 0.360^{o} C$ $= - 0.88 - 2.46$ $= - 3.34 F = - 3.34 \times 96500 J$ $= - 322.31 k J = - 322 k J$

Q. The rusting of iron takes place as follows RCOCl+H2​Pd/BaSO4​​RCHO+HCl BaSO4​ Calculate HCl for the net process.

l>II>lll>IV

IV>ll>lll>l

lll>IV>H>l

lll>ll>IV>l