Given the following equations, calculate the ΔH of reaction of ethylene gas (C2H4) with fluorine gas to make carbon tetrafluoride (CF4) gas and hydrogen fluoride gas. Show your work.
H2 (g) + F2 (g) → 2HF (g) ΔH = –537 kJ
C (s) + 2F2 (g) → CF4 (g) ΔH = –680 kJ
2C (s) + 2H2 (g) → C2H4 (g) ΔH = +52.3 kJ
Reverse eqn 3, and add 2x eqn 2 and 2x eqn 1 to get
C2H4 + 6F2 ==> 2CF4 + 4HF
To find the ΔH of the reaction of ethylene gas (C2H4) with fluorine gas to make carbon tetrafluoride (CF4) gas and hydrogen fluoride gas, we need to use the ΔH values of the given equations to calculate the overall ΔH.
Here's what we can do step by step:
1. Write down the given equations and their corresponding ΔH values:
i) H2 (g) + F2 (g) → 2HF (g) ΔH = –537 kJ
ii) C (s) + 2F2 (g) → CF4 (g) ΔH = –680 kJ
iii) 2C (s) + 2H2 (g) → C2H4 (g) ΔH = +52.3 kJ
2. Multiply all the equations to match the desired overall reaction:
As we want to calculate the ΔH for: C2H4 (g) + F2 (g) → CF4 (g) + HF (g)
i) Multiply equation i) by 2 to cancel out the 2HF (g):
2H2 (g) + 2F2 (g) → 4HF (g) ΔH = –1074 kJ
ii) Multiply equation ii) by 2 to cancel out the 2F2 (g):
2C (s) + 4F2 (g) → 2CF4 (g) ΔH = –1360 kJ
iii) No need to multiply equation iii) since it already has the desired compound.
3. Add up all the modified equations to obtain the overall balanced equation:
2C (s) + 4H2 (g) + 6F2 (g) → C2H4 (g) + 2CF4 (g) + 4HF (g) ΔH = –1074 kJ + –1360 kJ + +52.3 kJ
Simplifying the equation:
2C (s) + 4H2 (g) + 6F2 (g) → C2H4 (g) + 2CF4 (g) + 4HF (g) ΔH = –2381.7 kJ
Therefore, the ΔH of the reaction of ethylene gas (C2H4) with fluorine gas to make carbon tetrafluoride (CF4) gas and hydrogen fluoride gas is –2381.7 kJ.
To find the ΔH of the reaction of ethylene gas (C2H4) with fluorine gas to make carbon tetrafluoride (CF4) gas and hydrogen fluoride gas, we need to add the enthalpies of the individual reactions.
The given reactions are:
H2 (g) + F2 (g) → 2HF (g) ΔH = –537 kJ (Reaction 1)
C (s) + 2F2 (g) → CF4 (g) ΔH = –680 kJ (Reaction 2)
2C (s) + 2H2 (g) → C2H4 (g) ΔH = +52.3 kJ (Reaction 3)
We can rearrange Reaction 1 to match the stoichiometry in Reaction 3 by multiplying it by a factor of 2.
2(H2 (g) + F2 (g) → 2HF (g) ) ΔH = 2 * -537 kJ ΔH = -1074 kJ (Reaction 1)
Now, we can add Reactions 1, 2, and 3 to get the desired reaction:
2(H2 (g) + F2 (g)) + 2C (s) → 2HF (g) + CF4 (g) + C2H4 (g)
Adding the enthalpies of the individual reactions, we have:
ΔH = -1074 kJ + (-680 kJ) + (+52.3 kJ)
ΔH = -1074 kJ - 680 kJ + 52.3 kJ
ΔH = -1701.7 kJ
Therefore, the ΔH of the reaction of ethylene gas (C2H4) with fluorine gas is -1701.7 kJ.