A) \[-108.7\text{ }k\,cal\]
B) \[+108.7\text{ }k\,cal\]
C) \[-184.6\text{ }k\,cal\]
D) \[+184.6\text{ }k\,cal\]
Correct Answer: A
Solution :
\[P(s)+\frac{5}{2}C{{l}_{2}}(g)\xrightarrow{{}}PC{{l}_{5}}(s);\] \[\Delta H=?\] Given: \[2P(s)+2C{{l}_{2}}(g)\xrightarrow{{}}\] \[2PC{{l}_{3}}(l)+151.8\,kcal\,\,\,\,...(i)\] \[PC{{l}_{3}}(l)+C{{l}_{2}}(g)\xrightarrow{{}}\] \[PC{{l}_{5}}(s)+32.8\,kcal\,\,\,\,\,...(ii)\] Divide eq. (i) by 2 and add with equation (ii), we get \[P(s)+\frac{5}{2}C{{l}_{2}}(g)\xrightarrow{{}}PC{{l}_{5}}(s)\] \[+\left( \frac{151.8}{2}+32.8 \right)\,k\,cal\] \[\therefore \] \[\Delta {{H}_{f}}\]of \[PC{{l}_{5}}(s)=-\,\,(75.9+32.8)\] \[=-\,108.7\,k\,cal\]
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