Question

The heats of combustion of graphite and carbon monoxide, respectively are, $ -\text{ }393.5\text{ }kJ\text{ }mo{{l}^{-1}} $ and $ -283\text{ }kJ\text{ }mo{{l}^{-1}} $ . Therefore, the heat of formation of carbon monoxide in, $ kJ\text{ }mo{{l}^{-1}} $ is:

• A. 172.5
• B. $ -110.5 $
• C. $ -1070 $
• D. $ -676.5 $
• E. 110.5

Answer 1

Answer: (B)

Combustion of graphite is as follows:
$ C(s)+{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)+393.5\,kJ $ ...(i)
Combustion of carbon monoxide is, $ CO(g)+\frac{1}{2}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)+283\,kJ $ ...(ii)
For obtaining $ CO $ as a product we reverse the Eq. (ii)
as $ C{{O}_{2}}(g)\xrightarrow[{}]{{}}CO(g)+\frac{1}{2}{{O}_{2}}(g)-283\,kJ $ ...(iii)
$ C(g)+{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)+393.5\,kJ $ ...(iv)
On adding Eq. (i) and Eq. (iii)
$ C+\frac{1}{2}{{O}_{2}}\xrightarrow[{}]{{}}CO+(393.5-283) $
$ =+110.5\text{ }kJ $
$ \therefore $ Heat of formation of CO is
$ =+\text{ }110.5\text{ }kJ $
$ \therefore $ $ \Delta H=-110.5\text{ }kJ $