C2H4(g) + 3O2(g) => 2CO2(g) + 2H2O(l) ΔHf = -1411 kJ?
Data:
C2H4(g) + 3O2(g) => 2CO2(g) + 2H2O(l) ΔHf = -1411 kJ
C(s) + O2(g) => CO2(g) ΔHf = -393.5 kJ
H2(g) + ??O2(g) => H2O(l) ΔHf = -285.8 kJ
Use the data given above to find the standard enthalpy of formation of ethylene, C2H4(g).
ans-> 52 kJ/Mol
how show me steps please
Comments
C2H4(g) + 3O2(g) => 2CO2(g) + 2H2O(l) ΔHf = -1411 kJ
C(s) + O2(g) => CO2(g) ΔHf = -393.5 kJ
2H2(g) + O2(g) => 2H2O(l) ΔHf = -285.8 kJ
The formation of ethylene is 2C(s) + 2H2(g) -----> C2H4(g), so that's your end goal, to have that formula from adding these three equations. So first, that means we need the C2H4 on the other side, so we switch it around, and make the delta H positive.
2CO2(g) + 2H2O(l) => C2H4(g) + 3O2(g) ΔHf = +1411 kJ
In the final equation, there are two carbons, so we multiply the second equation by 2.
2C(s) + 2O2(g) => 2CO2(g) ΔHf = (-393.5 * 2) kJ
Now, all the O2 are gone, the C2H4 is on the right side, and the CO2 cancel out. So we can add them together now.
(2*-393.5 + 2*-285.8) - (-1411) = +52.4kJ/mol
First you have to write the reaction that they are talking about ... forming ethylene gas (C2H4) from its elements (C and H2) at 25 C.
2C(s) + 2H2(g) ==> C2H4(g)
Let me label the equations that were given.
(Equation 1) C2H4(g) + 3O2(g) => 2CO2(g) + 2H2O(l) ΔHf = -1411 kJ
(Equation 2) C(s) + O2(g) => CO2(g) ΔHf = -393.5 kJ
(Equation 3) H2(g) + 1/2 O2(g) => H2O(l) ΔHf = -285.8 kJ
Look at the desired result: 2C(s) + 2H2(g) ==> C2H4(g)
We need 2C(s) on the left side, so take Equation 2 and multiply it by 2.
2 x (Equation 2): 2C(s) + 2O2(g) ==> 2CO2(g) . . .ΔH = 2 x -393.5 kJ = -787.0 kJ
We also need 2H2(g) on the left side, so take Equation 3 and multiply by 2.
2 x (Equation 3): 2H2(g) + O2(g) ==> 2H2O . . .ΔH = 2 x -285.8 kJ = -571.6 kJ
We also need C2H4(g) on the right side, so reverse Equation 1.
-1 x (Equation): 2CO2(g) + 2H2O(l) ==> C2H4(g) + 3O2(g) . . .ΔH = -1 x -1411 kJ = +1411 kJ
Now add together the three equations that we modified. Note how H2O(l), CO2(g), and O2(g) cancel.
.2C(s) + 2O2(g) ==> 2CO2(g) . . . . . . . . . . . . ΔH = -787.0 kJ
+2H2(g) + O2(g) ==> 2H2O(l) . . . . . . . . . . . . .ΔH = -571.6 kJ
+2CO2(g) + 2H2O(l) ==> C2H4(g) + 3O2(g) . . .ΔH = +1411 kJ
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.2C(s) + 2H2(g) ==> C2H4(l) . . . . . . . . . . . . .ΔH = +52 kJ per mole of C2H4(l)
the answer is positive 1411kj period!!!!!