Ans: According to the question, heat of formation of CO and CO2 are as follows:
ΔHf(CO) = -110 kJ/mol
ΔHf(CO2) = -393 kJ/mol
The reaction is given as: CO(g) + 1/2 O2(g) → CO2(g)
For the heat of the reaction:
ΔH = ΔHf (products) - ΔHf (reactants)
= ΔHf(CO2) - [ΔHf(CO) + 1/2 ΔHf(O2)]
Substituting the values:
ΔH = -393kJ/mol - [ -110 + ½ X 0] ( the heat of formation of the elements in the standard states is 0.
ΔH = -393 kJ/mol - (-110 kJ/mol)
ΔH = -393 kJ/mol + 110 kJ/mol
ΔH = -283 kJ/mol
Therefore, the answer is -283kJ/mol
Enthalpy Change refers to the difference between the heat content of the initial and final state of the reaction. Change in enthalpy can prove to be of great importance to find whether the reaction is exothermic or endothermic.
dH = dU + d(PV)
The above equation can be written in the terms of initial and final states of the system which is defined below:
UF – UI = qP –p(VF – VI)
Or qP = (UF + pVF) – (UI + pVI)
Enthalpy (H) can be written as H= U + PV. Putting the value in the above equation, we obtained:
qP = HF – HI = ∆H
Hence, change in enthalpy ∆H = qP, referred to as the heat consumed at a constant pressure by the system. At constant pressure, we can also write,
∆H = ∆U + p∆V
To specify the standard enthalpy of any reaction, it is calculated when all the components participating in the reaction i.e., the reactants and the products are in their standard form. Therefore the standard enthalpy of reaction is the enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions.