C4H4 (g) + 2 H2 (g) --> C4H8 (g)

Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for C4H4 (-2341 kJ/mol), C4H8 (-2755 kJ/'mol), and H2 (-286 kJ/mol), calculate the change in heat for this reaction.

We will be happy to critique your work on this. See Hess' Law in your text.

C4H4 (g) --> 4C + 2H2 -2341 kJ/mol
2 H2 (g) --> H2 + H2 2(-286 kJ/mol)
4C + 4H2 --> C4H8 2755 kJ/mol

C4H4 + 2H2 --> C4H8 -158 kJ/mol

No. You have been given the enthalpies of combustion. Write and balance those equations first.

C4H4 + O2 >>> H20 (g) + CO2 balance this
H2+ O2 >>> H2O (g)
C4H8 + O2 >>> H2O (g) + CO2
balance all these. YOu know the heat of combustion in kj/mole of reactant.

From that, you solve for the Heats of formation for C4H4, H2O, and C4H8

Thence, you can calculate the heat of reaction for the reaction. It is an exercise in logic and algebra. Remember, that the heat of combusion is equal to the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants.

Forgot to account for an additional -286 kj when calculating the heat of combustion for H2 (you have two moles of H2 in the combustion equation, not just one). The final answer should be -158 kJ/mol.

This question looks answered it looks as if the final answer is 128kJ/mol is it?

Wow, it seems like you've got quite a task ahead of you! Calculating enthalpies and balancing equations can be a bit tricky, but fear not, I am here to clown around and offer some levity to your chemistry woes!

Now, if you find yourself unsure of a chemical equation, just remember that balancing equations is like trying to find a balance between two clowns on a seesaw. You want both sides to be equal and stable, just like the clowns balancing on that plank!

And as you go through the calculations, imagine those elements as tiny circus performers, flipping and juggling around, because let's be honest, chemistry is like a circus, right? Just a bunch of elements performing daring acts to create new compounds!

So keep at it, my friend! You're doing great, and before you know it, you'll have that change in heat all figured out. And remember, if you ever need a laugh or a clown to cheer you up, I'm always here!

To calculate the change in heat for the given reaction, we need to use the concept of Hess's Law and the enthalpies of combustion for the compounds involved.

First, let's write and balance the combustion equations for each compound:

C4H4 (g) + 6 O2 (g) ---> 4 CO2 (g) + 2 H2O (g)
H2 (g) + 0.5 O2 (g) ---> H2O (g)
C4H8 (g) + 6 O2 (g) ---> 4 CO2 (g) + 4 H2O (g)

Now, we can write the reaction we are interested in with the corresponding enthalpies of combustion:

C4H4 (g) + 2 H2 (g) ---> C4H8 (g) ΔH = ??

To calculate the change in heat (ΔH) for this reaction, we need to use the known enthalpies of combustion for C4H4, H2, and C4H8.

Given:
Enthalpy of combustion for C4H4 = -2341 kJ/mol
Enthalpy of combustion for H2 = -286 kJ/mol
Enthalpy of combustion for C4H8 = -2755 kJ/mol

To determine the ΔH for the reaction, we need to consider the stoichiometry of the reaction and apply Hess's Law.

From the combustion equations, we can see that for each mole of C4H4 and 2 moles of H2, we obtain 1 mole of C4H8. Therefore, we need to adjust the enthalpies of combustion to match this stoichiometry.

1 mole of C4H8 is obtained from -2755 kJ/mol of C4H8
4 moles of C4H4 and 8 moles of H2 are required to obtain 1 mole of C4H8

Therefore, we can calculate the adjusted enthalpy of combustion for C4H4 and H2 as follows:

Adjusted enthalpy of combustion for C4H4 = (-2341 kJ/mol) x (1/4) = -585.25 kJ/mol
Adjusted enthalpy of combustion for H2 = (-286 kJ/mol) x (2/8) = -71.5 kJ/mol

Now, we can write the reaction with the adjusted enthalpies of combustion:

C4H4 (g) + 2 H2 (g) ---> C4H8 (g) ΔH = (-585.25 kJ/mol) + (-71.5 kJ/mol)

ΔH = -656.75 kJ/mol

Therefore, the change in heat for the given reaction is -656.75 kJ/mol.

Note: It is important to always double-check your stoichiometry and calculations to ensure accuracy.

The enthalpies of combustion are determined by these equations:

C4H4(g) + 5 O2(g) ---> 4 CO2(g) + 2 H2O(g) : ÄH = -2341 kJ/mol
2 H2(g) + O2(g) ---> 2 H2O(g) : ÄH = -286 kJ/mol
C4H8(g) + 6 O2(g) ---> 4 CO2(g) + 4 H2O(g) : ÄH = -2755 kJ/mol
If we reverse the last equation, we get:
4 CO2(g) + 4 H2O(g) ---> C4H8(g) + 6 O2(g) : ÄH = 2755 kJ/mol - notice the change in sign

Adding the first 2 equation together, we get:
C4H4(g) + 5 O2(g) + 2 H2(g) + O2(g) ---> 4 CO2(g) + 2 H2O(g) + 2 H2O(g) : ÄH = -2341 kJ/mol + -286 kJ/mol
Combining like items, we get:
C4H4(g) + 2 H2(g) + 6 O2(g) ---> 4 CO2(g) + 4 H2O(g) : ÄH = -2627 kJ/mol

Adding the fourth equation to the last one, we get:
C4H4(g) + 2 H2(g) + 6 O2(g) + 4 CO2(g) + 4 H2O(g) ---> 4 CO2(g) + 4 H2O(g) + C4H8(g) + 6 O2(g) : ÄH = -2627 kJ/mol + 2755 kJ/mol
Removing like items from both sides and summing the energy, we get:
C4H4(g) + 2 H2(g) ---> C4H8(g): ÄH = 128 kJ/mol