2C2H6 + 7O2 yields 4CO2 + 6H2O
What is the (∆rH°) of this equation?
I think I was able to calculate 3119.656 KJ, but is that correct?
Let's put it this way. I don't want to find my tables, look up the values, find my calculator, plug in the numbers and type in those numbers. However, if you used
dHo rxn = ((n*dHo products) - (n*dH reactants) your answer should be good.
To find the standard enthalpy change (∆rH°) for the given equation, you need to use a set of known standard enthalpy values for the reactants and products involved in the reaction.
The standard enthalpy change (∆rH°) can be calculated using the formula:
∆rH° = ∑(∆H°f, products) - ∑(∆H°f, reactants)
Where:
- ∆rH° is the standard enthalpy change of the reaction
- ∆H°f, products is the standard enthalpy of formation for the products
- ∆H°f, reactants is the standard enthalpy of formation for the reactants
First, you need to know the standard enthalpy of formation (∆H°f) values for all the compounds involved in the reaction. These values represent the enthalpy change when one mole of a substance is formed from its constituent elements in their standard states at standard conditions (1 atm and 25°C).
Here are the standard enthalpy of formation values for the compounds involved:
- ∆H°f(CO2) = -393.5 kJ/mol
- ∆H°f(H2O) = -285.8 kJ/mol
- ∆H°f(C2H6) = -84.7 kJ/mol
Using these values, we can now calculate ∆rH°:
∆rH° = (4 * ∆H°f(CO2)) + (6 * ∆H°f(H2O)) - (2 * ∆H°f(C2H6))
∆rH° = (4 * -393.5 kJ/mol) + (6 * -285.8 kJ/mol) - (2 * -84.7 kJ/mol)
∆rH° = -1574 kJ/mol + (-1714.8 kJ/mol) + 169.4 kJ/mol
∆rH° = -2119.4 kJ/mol
Therefore, the standard enthalpy change (∆rH°) of the given reaction is -2119.4 kJ/mol.