A) \[-108k,\,cal\]
B) \[-196k,\,cal\]
C) \[-98k,\,cal\]
D) \[54k,\,cal\]
Correct Answer: C
Solution :
\[{{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow{{}}2HCl(g)\] \[\Delta H=-44k.\,cal\] ??(i) \[2Na(s)+2HCl(g)\xrightarrow{{}}2NaCl(s)+{{H}_{2}}(g)\] \[\Delta H=-152k.cal\] ??.(ii) \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{{}}NaCl\,(s)\,\Delta H=?\] By adding eq. (i) and (ii) we have \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\to NaCl\,\Delta H=-98k.cal\] \[2Na(s)+C{{l}_{2}}(g)\xrightarrow{{}}2NaCl,\] \[\Delta H=-196k.\,\,cal\] ?..(iii) dividing eq. (iii) by 2 we have \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\to NaCl\Delta H=-98k.cal\]You need to login to perform this action.
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