Use the information below to answer this question
C(s) + O2(g) -> CO2(g) ΔH = -394 kJ/mol
H2(g) + 1/2 O2(g) -> H2O(l) ΔH = -286 kJ/mol
4C(s) + 5H2(g) -> C4H10 (g) ΔH = -126 kJ/mol
The standard enthalpy of combustion of butane, in kJ/mol, is:
A -2880
B -2590
C -806
D -554
Could someone show me how to do this question? Thank you!
eqn 3 reversed + 4*eqn 1 + 5*eqn 2
Post your work if you get stuck.
So the answer is A, right? Thanks for the help!
I didn't do the math.
Yes, A is correct.
Hey! That's pretty goooood!
To find the standard enthalpy of combustion of butane, we need to use the given thermochemical equations and apply Hess's Law. Hess's Law states that if a reaction can be expressed as the sum of two or more other reactions, then the change in enthalpy for the overall reaction will be the sum of the enthalpy changes of the individual reactions.
First, let's write the balanced chemical equation for the combustion of butane (C4H10):
C4H10(g) + 13/2 O2(g) -> 4CO2(g) + 5H2O(l)
Now, we need to find a combination of the given reactions that will give us this overall reaction.
We can see that the first reaction already has CO2 as a product, so we can use it as it is. However, it is in the wrong ratio, so we need to multiply it by 4:
4C(s) + 4O2(g) -> 4CO2(g) (multiply by 4)
Next, we see that the second reaction has water (H2O) as a product, but it is in the wrong ratio, so we need to multiply it by 5:
5H2(g) + 5/2 O2(g) -> 5H2O(l) (multiply by 5)
Now, we need to balance the oxygen atoms on both sides of the equation. In the first equation, we have 4 oxygen atoms on the left side but 8 on the right side. In the second equation, we have 5/2 oxygen atoms on the left side but 5 oxygen atoms on the right side. We can multiply the second equation by 4 to make the number of oxygen atoms the same in both equations:
20H2(g) + 10O2(g) -> 20H2O(l) (multiply by 4)
Now, we have the necessary reactions to form the combustion of butane equation. We can add these reactions together:
4C(s) + 4O2(g) + 20H2(g) + 10O2(g) -> 4CO2(g) + 5H2O(l) + 20H2O(l)
Simplifying the equation:
4C(s) + 24H2(g) + 14O2(g) -> 4CO2(g) + 25H2O(l)
We can see that we have 4 moles of carbon (C), so we need to multiply the enthalpy change of the first reaction by 4:
4 ΔH1 = 4 * (-394 kJ/mol) = -1576 kJ/mol
We also have 25 moles of water (H2O), so we need to multiply the enthalpy change of the second reaction by 25:
25 ΔH2 = 25 * (-286 kJ/mol) = -7150 kJ/mol
Finally, we can calculate the enthalpy change for the combustion of butane by adding these two values together:
ΔH(combustion of butane) = ΔH1 + ΔH2 = -1576 kJ/mol - 7150 kJ/mol = -8726 kJ/mol
However, the question asks for the standard enthalpy of combustion, so we need to divide this value by the coefficient in front of butane (C4H10), which is 1:
Standard ΔH(combustion of butane) = -8726 kJ/mol / 1 = -8726 kJ/mol
Therefore, the standard enthalpy of combustion of butane is -8726 kJ/mol.
Looking at the given answer choices, none of them match the calculated value of -8726 kJ/mol, so there may be an error or omission in the provided information.