Calculate the enthalpy change for the reaction 2C + H2 yield C2H2 given the following reactions and their respective enthalpy changes:
C2H2 + 5/2 O2 yield 2CO + H2O -1299.6
C + O yield CO2 -393.5
H2 + 1/2 O2 yield H2O -285.9
I don't even know how to start and the example in the book doesn't explain anything!!!!
Look in your text about Hess' Law. That is what this problem is. The idea is to multiply (if necessary) an equation AND/OR reverse it so that all of them added together will give you the final equation you want.
If you get stuck let us know what you don't understand (be specific---; i.e., I don't know where to start or I don't have a clue doesn't help us determine the problem you are having.)
Check you post carefully. I don't think all of the equations you have written are balanced. For example, look at the ones with CO. Are those CO2?
To calculate the enthalpy change for the reaction 2C + H2 → C2H2, you can use Hess's law, which states that the enthalpy change for a reaction is the sum of the enthalpy changes of the reactions that provide an indirect path from the reactants to the products.
Here's a step-by-step approach to solve the problem:
Step 1: Write the given reactions
Write the given reactions and their respective enthalpy changes:
1. C2H2 + 5/2 O2 → 2CO + H2O ΔH1 = -1299.6 kJ/mol
2. C + O → CO2 ΔH2 = -393.5 kJ/mol
3. H2 + 1/2 O2 → H2O ΔH3 = -285.9 kJ/mol
Step 2: Manipulate the equations
Multiply the second equation by 2 to balance the carbon atoms:
4. 2C + 2O → 2CO2 ΔH4 = -2 * (-393.5 kJ/mol) = +787.0 kJ/mol
Step 3: Combine the equations
Now, we need to combine the equations in a way that cancels out the reactants and products that appear on both sides of the equation.
Multiply equation 1 by 2 to balance the carbon atoms:
5. 2C2H2 + 5O2 → 4CO + 2H2O 2 * ΔH1 = -2 * (-1299.6 kJ/mol) = +2599.2 kJ/mol
Multiply equation 3 by 2 to balance the hydrogen atoms:
6. 2H2 + O2 → 2H2O 2 * ΔH3 = -2 * (-285.9 kJ/mol) = +571.8 kJ/mol
Now, add equations 4, 5, and 6:
2C + H2 + 5/2 O2 + 2H2 + O2 → C2H2 + 4CO + 2H2O
ΔH = ΔH4 + ΔH1 + ΔH3
= +787.0 kJ/mol + 2599.2 kJ/mol + 571.8 kJ/mol
= +3957.8 kJ/mol
Therefore, the enthalpy change for the reaction 2C + H2 → C2H2 is +3957.8 kJ/mol.