Calculate the enthalpy change for:
C2H4(g) + H2 --> C2H6(g)
delta Hrxn:
H2(g) + 1/2O2 --> H2O (l)
C2H4(g) + 3O2 --> 2H20(l) + 2CO2(g)
C2H6(g) + 7/2O2(g) --> 3H20(l) + 2CO2(g)
I know I have to flip the third reaction but I don't know what to do next
Add equation 1 and equation 2 to the reverse of equation 3 and cancel molecules that appear on both sides of the equation to end up with the equation you want. For the dH rxn in this case add dH for 1 to dH for 2 to the negative dH for 3.
@k8Lbo3
To calculate the enthalpy change for the reaction:
C2H4(g) + H2 --> C2H6(g)
You need to consider the enthalpy changes of the individual reactions and apply them accordingly. Here are the steps to follow:
1. Write down the balanced equation for the given reaction and assign the corresponding enthalpy change (ΔH) values to each compound:
C2H4(g) + H2(g) --> C2H6(g)
Now, let's assign the ΔH values:
ΔH1 = ΔHf(C2H6) - ΔHf(C2H4) - ΔHf(H2)
2. Flip the third reaction:
- C2H6(g) + 7/2O2(g) --> 3H2O(l) + 2CO2(g)
3. Multiply the equation by a factor to relate the number of moles of hydrogen gas:
- 2C2H6(g) + 7O2(g) --> 6H2O(l) + 4CO2(g)
Now, let's assign the ΔH values:
ΔH2 = ΔHf(2CO2) + ΔHf(6H2O) - ΔHf(2C2H6)
4. Multiply the second equation by a factor to relate the number of moles of ethylene (C2H4):
- C2H4(g) + 3O2(g) --> 2H2O(l) + 2CO2(g)
Now, let's assign the ΔH values:
ΔH3 = ΔHf(2CO2) + ΔHf(2H2O) - ΔHf(C2H4)
5. Add up the equations and their respective ΔH values:
ΔHrxn = ΔH1 + ΔH2 - ΔH3
Remember to use the enthalpy of formation values (ΔHf) for each compound, which represent the enthalpy change when 1 mole of the compound is formed from its elements in their standard state at the reference temperature.
Note: Make sure to use the correct ΔHf values and be consistent with units (kJ/mol or kJ).
To calculate the enthalpy change for a chemical reaction, you can use Hess's Law. In this case, you need to find a combination of the given reactions that can be manipulated to yield the desired reaction. Here's how you can proceed:
1. Start with the given reactions and their respective enthalpy changes:
a. H2(g) + 1/2O2(g) → H2O(l) (ΔH1)
b. C2H4(g) + 3O2(g) → 2H2O(l) + 2CO2(g) (ΔH2)
c. C2H6(g) + 7/2O2(g) → 3H2O(l) + 2CO2(g) (ΔH3)
2. You correctly noted that the third reaction needs to be flipped. When you flip a reaction, the sign of the enthalpy change also changes. So, the new reaction becomes:
C2H6(g) + 7/2O2(g) ← 3H2O(l) + 2CO2(g) (ΔH3 flipped)
3. Next, you need to manipulate the given reactions to make them align with the desired reaction equation. To achieve this, you can multiply the reactions by appropriate coefficients so that the newly formed equations add up to the desired reaction equation. Here's what you can do:
a. 2 × (H2(g) + 1/2O2(g) → H2O(l)) = 2H2(g) + O2(g) → 2H2O(l) (ΔH1 doubled)
b. 3/2 × (C2H6(g) + 7/2O2(g) ← 3H2O(l) + 2CO2(g)) = (3/2)C2H6(g) + (21/4)O2(g) ← (9/2)H2O(l) + 3CO2(g) (ΔH3 flipped and scaled)
4. Now, sum up the manipulated reactions to get the desired reaction equation:
2H2(g) + O2(g) + (3/2)C2H6(g) + (21/4)O2(g) → 2H2O(l) + 3CO2(g) (Combined equation)
5. Finally, add up the enthalpy changes of the manipulated reactions to get the overall enthalpy change for the desired reaction:
ΔHrxn = (ΔH1 doubled) + (ΔH2) + (ΔH3 flipped and scaled)
Note: Make sure to adjust the coefficients in front of the reactions when calculating the enthalpy changes.
I hope this explanation helps you understand how to calculate the enthalpy change using Hess's Law.