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Question: Calculate the enthalpy of formation for ethylene from the following data: 1\. \(C(graphite) + {O_{...

Calculate the enthalpy of formation for ethylene from the following data:
1. C(graphite)+O2(g)→CO2(g)C(graphite) + {O_{2(g)}} \to C{O_{2(g)}} ; ΔH=−393.5kJ\Delta H = - 393.5kJ
2. H2(g)+12O2(g)→H2O(l){H_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to {H_2}{O_{(l)}} ; ΔH=286.2kJ\Delta H = 286.2kJ
3. C2H4(g)+3O2(g)→2CO2(g)+2H2O(l){C_2}{H_{4(g)}} + 3{O_{2(g)}} \to 2C{O_{2(g)}} + 2{H_2}{O_{(l)}} ; ΔH=−1410.8kJ\Delta H = - 1410.8kJ

A. 54.1kJ54.1kJ
B. 44.8kJ44.8kJ
C. 51.4kJ51.4kJ
D. 48.4kJ48.4kJ

Explanation

Solution

Since enthalpy follows scalar addition and subtraction, use these reactions and add and subtract them is such a way that the net reaction is the reaction for formation of ethylene. Then use hess’s law to get the enthalpy value

Complete step by step answer:
The reaction for the formation of ethylene gas would be:
2C(graphite)+2H2(g)→C2H4(g)2C(graphite) + 2{H_{2(g)}} \to {C_2}{H_{4(g)}}
Hence, we need to find out how to get this possible reaction by adding and subtracting the above equations.
C(graphite)+O2(g)→CO2(g)C(graphite) + {O_{2(g)}} \to C{O_{2(g)}} …………... Equation A
H2(g)+12O2(g)→H2O(l){H_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to {H_2}{O_{(l)}} ……………………...Equation B
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l){C_2}{H_{4(g)}} + 3{O_{2(g)}} \to 2C{O_{2(g)}} + 2{H_2}{O_{(l)}}...................Equation C

If we multiply equation B by two, we get:
2H2(g)+O2(g)→2H2O(l)2{H_{2(g)}} + {O_{2(g)}} \to 2{H_2}{O_{(l)}} and the value for ΔH=−286.2kJ\Delta H = - 286.2kJ also changed by a factor of 2
So the new ΔH=−572.4kJ\Delta H = - 572.4kJ. Let us name this equation as equation D

Let us also multiply equation A also by two, we get:
2C(graphite)+2O2(g)→2CO2(g)2C(graphite) + 2{O_{2(g)}} \to 2C{O_{2(g)}} and the value for ΔH=−393.5kJ\Delta H = - 393.5kJ also be changed by a factor of 2, so the new ΔH=−787kJ\Delta H = - 787kJ. Let us name this equation E

Now let us add equation D and equation E
We get
2H2(g)+2O2(g)+C(graphite)+O2(g)→CO2(g)+2H2O(l)2{H_{2(g)}} + 2{O_{2(g)}} + C(graphite) + {O_{2(g)}} \to C{O_{2(g)}} + 2{H_2}{O_{(l)}}

Now, if we add subtract reverse of equation C we get:
H2+2O2+C(graphite)+O2+CO2+H2O→C2H4+O2+CO2+H2O{H_2} + 2{O_2} + C(graphite) + {O_2} + C{O_2} + {H_2}O \to {C_2}{H_4} + {O_2} + C{O_2} + {H_2}O
Cancelling out all the common terms we get the final value as:
2C(graphite)+2H2(g)→C2H4(g)2C(graphite) + 2{H_{2(g)}} \to {C_2}{H_{4(g)}}
Since enthalpy is a state function we can add the values required.

ΔH=2ΔHA+2ΔHB−ΔHC\Delta H = 2\Delta {H_A} + 2\Delta {H_B} - \Delta {H_C}
So we know:
ΔHA=−747kJ\Delta {H_A} = - 747kJ
ΔHB=−572.4kJ\Delta {H_B} = - 572.4kJ
ΔHC=−1410.8kJ\Delta {H_C} = - 1410.8kJ
Substituting These values in the above equation, we get:
−747−572.4−(−1410.8)- 747 - 572.4 - ( - 1410.8)
Solving this equation we get:
ΔHf=51.4kJ\Delta {H_f} = 51.4kJ

So, the correct answer is Option C.

Note: Enthalpy is a state function which means it is independent of the path followed by the process. Hess’s law is based on the law of conservation of energy. It allows you to get the value of enthalpy formation without actually calculating it.