Question

The heat of formation of $PC{{l}_{5}}(s)$ will be: Given: $2P(s)+3C{{l}_{2}}(g)\xrightarrow{{}}2PC{{l}_{3}}(l);$ $\Delta H=-151.8\,k\,cal$ $PC{{l}_{3}}(l)+C{{l}_{2}}(g)\xrightarrow{{}}PC{{l}_{5}}(s)\,;$ $\Delta H=-\,32.8\,k\,cal$

• A. $-108.7\text{ }k\,cal$
• B. $+108.7\text{ }k\,cal$
• C. $-184.6\text{ }k\,cal$
• D. $+184.6\text{ }k\,cal$

Answer 1

Answer: (A)

$-108.7\text{ }k\,cal$

Answer 2

$P(s)+\frac{5}{2}C{{l}_{2}}(g)\xrightarrow{{}}PC{{l}_{5}}(s);$ $\Delta H=?$ 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 e (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 $PC{{l}_{5}}(s)=-\,\,(75.9+32.8)$ $=-\,108.7\,k\,cal$