Standard heat of formation of caebonmonooxide at 298 K is -110.5 kJ. Calculate the standard heat of formation of carbon mono oxide at constant volume and 298 K.

Explanation

Heat of formation of CO (g) is represented in the form of following thermochemical equation :
`C(s)+(1)/(2)O_(2)(g)rarr CO(g) , Delta H=-110.5 kJ`
`Delta n` for this equation = Number of gaseous moles of product - Number of gaseous mole of reactant
`Delta n_(g)=1-(1)/(2)=+(1)/(2) , R =8.314xx10^(-2)` kJ/K/mol., T = 298 K
Now `Delta H=Delta E+Delta n_(g)RT`
`Delta E` = Standard heat of reaction at constant volume = ?
Thus, `-110.5 = Delta E+(+(1)/(2))xx8.314xx10^(-3)xx298`
`Delta E =(-110.5 -(1)/(2)xx8.314xx10^(-3)xx298)kJ//mol. =-(-110.5-1.238)kJ//mol. =-111.738 kJ//mol`.