Question

Given the thermochemical equation C + O2 → CO2 ΔH = -393 kJ what

is the ΔH for this reaction? Please explicitly include the sign. 2C
+ 2O2 → 2CO2

Answer 1

The ΔH for the reaction 2C + 2O2 → 2CO2 is -786 kJ, as it is double the ΔH of the given reaction with carbon and oxygen forming carbon dioxide which has ΔH = -393 kJ.

Step-by-step explanation:

The given thermochemical equation for the reaction of carbon with oxygen to form carbon dioxide is C + O2 → CO2 ΔH = -393 kJ. For the reaction 2C + 2O2 → 2CO2, we can deduce that its enthalpy change (ΔH) would simply be double the enthalpy change of the given reaction since the coefficients of all reactants and products have been multiplied by 2.

Using the principle that if a chemical reaction is reversed, the sign on ΔH is changed and if a multiple of a chemical reaction is taken, the same multiple of the ΔH is taken as well, we find that the enthalpy change for the reaction 2C + 2O2 → 2CO2 is ΔH = 2 × -393 kJ = -786 kJ.

The equation clearly shows that the reaction is exothermic, as indicated by the negative sign for ΔH, indicating energy is released.

Answer 2

Doubling the chemical equation also doubles the enthalpy change (∆H), so for the reaction 2C + 2O₂ → 2CO₂, the ∆H is -786 kJ.

Step-by-step explanation:

When doubling the coefficients in a thermochemical equation, you also double the ∆H value. Accordingly, the reaction 2C + 2O₂ → 2CO₂ has a ∆H of -786 kJ, since the original reaction C + O₂ → CO₂ has a ∆H of -393 kJ. When the equation is multiplied by two, the enthalpy change (∆H) will also be multiplied by two to maintain the correlation between the quantity of reactants and products and the energy change. Therefore, 2C + 2O₂ → 2CO₂ has a ∆H of -393 kJ x 2, which equals -786 kJ.