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)

$ 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 $