Q. Mark out the enthalphy of formation of carbon monoxide (CO) Given, $C(s) + O_{2}(g) \rightarrow CO(g),\ \mathrm{\Delta}H = - 393.3$ kJ/mol $CO(g) + \frac{1}{2}O_{2}(g) \rightarrow {CO}_{2}(g),\ \mathrm{\Delta}H = - 282.8\ kJ/mol$
A) 110.5 kJ/mol
B) 676.1 kJ/mol
C) 282.8 kJ/mol
D) 300.0 kJ/mol
Explanation
(a)
$C(s) + \frac{1}{2}O_{2}(g) \rightarrow CO(g)$
This equation can be obtained by subtraction of
$\left\lbrack CO(g) + \frac{1}{2}O_{2}(g) \rightarrow {CO}_{2}(g) \right\rbrack$
from $\lbrack C(s) + \frac{1}{2}O_{2}(g) \rightarrow {CO}_{2}(g)\rbrack$
Hence, $\mathrm{\Delta}H_{f} = (CO) = \lbrack - 393.3 - ( - 282.8)\rbrack$ kJ
$=$110.5 kJ/mol
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.