Question

Calculate the standard enthalpy of formation (∆H°f) at 298.15 K for CO(g)     

Known: Reaction      ∆H°R298.15 K, cal/mol    for C(gr) + O2(g) → CO2(g)  =     -94,052 cal/mol and  2CO(g) + O2(g) → 2CO2(g)     = -135,272

Answer 1

Calculate the standard enthalpy of formation (∆H°f) at 298.15 K for CO(g)    

Known: Reaction      ∆H°R298.15 K, cal/mol for C(gr) + O2(g) → CO2(g) = -94,052 cal/mol

Delta H^oreaction=∑Delta H of (products)−∑ Delta H of (Reactants)

CO2(g) + H2(g)→CO(g) + H2O(g).

Here enthalpy of reactant is given i.e -94,052 cal/mol

Delta H^o = [1×enthalpy of CO + ethalpy of H2O] – [1×enthalpy of H + Ethalpy of CO2]

Standard enthalpy of water = −285.8 KJ/mol

Standard enthalpy of CO = −110.53 KJ/mol

Standard enthalpy of H2= 0 KJ/mol

Standard enthalpy of CO2 (given)= -94,052 cal/mol = -94.052 Kcal/mol = -393.513 KJ/mol

Delta H^o = [ 1×−110.53 + 1× −285.8] – [ 1×-393.513 + 1× 0] = -2.877 KJ/mol

(Delta H^o (formation of CO) = -2.877 KJ/mol)