Using Hess’s Law, determine the enthalpy of reaction for,C2H4(g) + H2(g) –> C2H6(g) ΔH = ?Using the following reactions:C2H4(g) + 3O2(g) –> 2CO2(g) + 2H2O(l) ΔH = -1401 kJC2H6(g) + 7/2O2(g) –> 2CO2(g) + 3H2O(l) ΔH = -1550 kJH2(g) + 1/2O2(g) –> H2O(l) ΔH = -286 kJGroup of answer choices243 kJ-311 kJ95 kJ-137 kJ
Question
Using Hess’s Law, determine the enthalpy of reaction for,C2H4(g) + H2(g) –> C2H6(g) ΔH = ?Using the following reactions:C2H4(g) + 3O2(g) –> 2CO2(g) + 2H2O(l) ΔH = -1401 kJC2H6(g) + 7/2O2(g) –> 2CO2(g) + 3H2O(l) ΔH = -1550 kJH2(g) + 1/2O2(g) –> H2O(l) ΔH = -286 kJGroup of answer choices243 kJ-311 kJ95 kJ-137 kJ
Solution
To solve this problem, we need to manipulate the given reactions in such a way that they add up to the desired reaction.
The desired reaction is: C2H4(g) + H2(g) –> C2H6(g)
The given reactions are:
- C2H4(g) + 3O2(g) –> 2CO2(g) + 2H2O(l) ΔH = -1401 kJ
- C2H6(g) + 7/2O2(g) –> 2CO2(g) + 3H2O(l) ΔH = -1550 kJ
- H2(g) + 1/2O2(g) –> H2O(l) ΔH = -286 kJ
We can see that the second reaction is the reverse of the desired reaction, so we will reverse it and change the sign of ΔH:
2CO2(g) + 3H2O(l) –> C2H6(g) + 7/2O2(g) ΔH = 1550 kJ
Now, we need to cancel out the extra substances. We can do this by adding the first and third reactions:
C2H4(g) + 3O2(g) + H2(g) + 1/2O2(g) –> 2CO2(g) + 2H2O(l) + H2O(l) ΔH = -1401 kJ - 286 kJ = -1687 kJ
This gives us:
C2H4(g) + H2(g) + 7/2O2(g) –> 2CO2(g) + 3H2O(l) ΔH = -1687 kJ
Now, we can add this reaction to the reversed second reaction:
C2H4(g) + H2(g) –> C2H6(g) ΔH = 1550 kJ - 1687 kJ = -137 kJ
So, the enthalpy of the reaction C2H4(g) + H2(g) –> C2H6(g) is -137 kJ.
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