Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2CO(g) + 2NO(g)2CO2(g) + N2(g)

Answer 1

Ignore N₂(g) because that is the natural state of the element (ΔH_f=0)

ΔH_rxn = 2ΔH_f,CO2 - 2ΔH_f,NO - 2ΔH_f,CO

Consider a tutor.

Answer 2

Hi Jaden!

To calculate the standard enthalpy change for a reaction, we just need the chemical equation for the reaction and the heats of formation for each species (reactant and product) involved in the reaction.

After we have all of those, we can find the standard enthalpy change of the reaction by calculating ∑Enthalpy of Products - ∑Enthalpy of Reactants. In other words, we need to find the total enthalpy of formation for the products and subtract from that the total enthalpy of formation for the reactants. Quick note though: when you look up a table value for heat of formation, that's almost always in kJ/mol--so if your reaction has more than 1 mole of a species, you need to take that into account. For example, if you have a reaction that forms 2 moles of water, you need to multiply the heat of formation of water by 2 to account for that 2 moles.

For this question, our equation is 2 CO_(g) + 2 NO_(g) --> 2 CO_{2 (g)} + N_{2 (g)}. Now, let's find the heat of formation for each of those substances, using a reference table in a textbook or found somewhere else online (I used this one: https://www.thoughtco.com/common-compound-heat-of-formation-table-609253)

CO_(g) -110.5 kJ/mol

NO_(g) 90.4 kJ/mol

CO_{2 (g)} -393.5 kJ/mol

N_{2 (g)} 0 (this isn't in the reference table, but this is the natural state that Nitrogen is found in at standard conditions, diatomic gas. The heat of formation of elements in their natural state is 0, so it's 0 here).

The sum of product enthalpies, based on the chemical equation, is then 2(-393.5)+0 = -787 kJ/mol

The sum of reactant enthalpies, based on the chemical equation, is 2(-110.5) + 2(90.4) = -40.2 kJ/mol.

Therefore, the standard enthalpy change of this reaction is -787 kJ/mol-(-40.2 kJ/mol) = -746.8 kJ/mol.