
- A

- ✓

- C

- D







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$\frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = k\left[ {{N_2}{O_5}} \right]$ ,
$\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = k'\left[ {{N_2}{O_5}} \right]$,
$\frac{{d\left[ {{O_2}} \right]}}{{dt}} = k''\left[ {{N_2}{O_5}} \right]$
The relationship between $k$ and $k'$ and between $k$ and $k^{"}$ are
(Assume $100\%$ ionization)
$A.$ $0.500\,M\,C _2 H _5 OH ( aq )$ and $0.25\, M\, KBr ( aq )$
$B.$ $0.100\,M\,K _4\left[ Fe ( CN )_6\right]$ (aq) and $0.100\, M$ $FeSO _4\left( NH _4\right)_2 SO _4$ (aq)
$C.$ $0.05 \,M\, K _4\left[ Fe ( CN )_6\right]( aq )$ and $0.25\, M\, NaCl$ (aq)
$D.$ $0.15\, M\, NaCl ( aq )$ and $0.1\, M BaCl _2$ (aq)
$E.$ $0.02\, M\, KCl\, MgCl _{2 .} 6 H _2 O ( aq )$ and $0.05\, M$ $KCl ( aq )$

(At. nos. $Ce = 58, Sm = 62,$$ Eu = 63, Yb = 70$)