A) \[-108\,k.cal\]
B) \[-196\,k.cal\]
C) \[-98\,k.cal\]
D) \[54\,k.cal\]
Correct Answer: C
Solution :
\[{{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow{{}}2HCl(g)\] \[\Delta H =-44\,k.cal\] ?(i) \[2Na(s)+2HCl(g)\xrightarrow{{}}2NaCl(s)+{{H}_{2}}(g)\] \[\Delta H=-152\,k.cal\] ?(ii) \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{{}}NaCl(s)\Delta H=?\] By adding eq. (i) and (ii) we have \[2Na(s)+C{{l}_{2}}(g)\xrightarrow{{}}2NaCl,\] \[\Delta H=-196\,k.cal\] ?(iii) \[=\frac{1}{2}M{{\upsilon }^{2}}+\frac{1}{2}M{{R}^{2}}\left( \frac{{{\upsilon }^{2}}}{{{R}^{2}}} \right)=M{{\upsilon }^{2}}\] So \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{1}{2}\] Given: Here, viscosity \[\eta =18\times {{10}^{-5}}\]poise\[=18\times {{10}^{-6}}kg/m\,\sec \] Radius \[r=0.3\,mm=0.3\times {{10}^{-3}}m\] Velocity \[\upsilon =1\,m/s\] So, \[F=6\times 3.14\times 18\times {{10}^{-6}}\times 0.3\times {{10}^{-3}}\times 1\] \[=101.73\times {{10}^{-9}}N\] dividing eq. (iii) by 2 we have \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\to NaCl\Delta H=-98\,k.cal\]
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