MCQ
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below :

$\frac{1}{2}C{l_2}(g)\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}Cl(g)\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}$ $C{l^ - }(g)\xrightarrow{{{\Delta _{Hyd}}{H^\Theta }}}C{l^ - }(aq)$

(using the data, 

${{\Delta _{diss}}H_{C{l_2}}^\Theta } = 240\,kJ\,mol^{-1}, {{\Delta _{eg}}H_{C{l}}^\Theta }= -349 \,kJ\,mol^{-1},$${{\Delta _{Hyd}}H_{C{l}}^\Theta }= -381 \,kJ\,mol^{-1}$ ) will be ............. $\mathrm{kJ\,mol}^{-1}$

  • A
    $+ 152$
  • $- 610$
  • C
    $- 850$
  • D
    $+ 120$

Answer

Correct option: B.
$- 610$
b
The energy involved in the conversion of

$\frac{1}{2} C_{2}(g)$ to $C l^{-1}(a q)$ is given by

$\Delta H=\frac{1}{2} \Delta_{d i s s} H_{C l_{2}}^{(-)}+\Delta_{e g} H_{C l}^{(-)}+\Delta_{h y l} H_{C l}^{(-)}$

Substituting various values from given data, we get

$\Delta H=\left(\frac{1}{2} \times 240\right)+(-349)+(-381) \,k J \,m o l^{-1}$

$=(120-349-381)\, k J\, m o l^{-1}$

$=-610 \,k J \,m o l^{-1}$

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