calculate the delta h for the reaction 2C(solid) + H2(gas)-->C2H2(gas) use the i
ID: 789124 • Letter: C
Question
calculate the delta h for the reaction 2C(solid) + H2(gas)-->C2H2(gas)
use the information below to manipulate a combination of 2 or more (if possible) equation then do the same to the delta H's
please use great detail and if i can completly understand every thing i will send extra points if possible
you must use
C2H5(g) + 5/2O2(g)--> 2CO2(g) + H2O(l) delta H =- 1299.6KJ/mole
C(s) + O2(g)---> CO2(g) delta H= -393.5KJ/mole
H2(g) + 1/2O2(g) ---> delta H=-285.8KJ/mole
Explanation / Answer
Reaction 1:
C2H5(g) + 5/2O2(g)--> 2CO2(g) + H2O(l) delta H =- 1299.6KJ/mole
Reaction 2:
C(s) + O2(g)---> CO2(g) delta H= -393.5KJ/mole
Reaction 3:
H2(g) + 1/2O2(g) ---> delta H=-285.8KJ/mole
first, lets use reaction to as our base reaction to get the equation,2C(solid) + H2(gas)-->C2H2(gas)
C(s) + O2(g)---> CO2(g) delta H= -393.5KJ/mole
as the final reaction contains two moles of C ,we multiply 2 with this equation
=>2C(s) + 2O2(g)---> 2CO2(g) delta H= -787KJ/mole
now adding reaction ,reaction 3 with this ,we get
2C(s) + 2O2(g)---> 2CO2(g) delta H= -787KJ/mole
+
H2(g) + 1/2O2(g) --->H2O(l) delta H=-285.8KJ/mole
----------------------------------------------------------------------------
2C(s) +H2(g) + 5/2O2(g)---> 2CO2(g)+H2O(l) delta H=-1072.8kJ
now, again adding reaction 1 to the above resulted equation,
2C(s) +H2(g) + 5/2O2(g)---> 2CO2(g)+H2O(l) delta H=-1072.8kJ
-
C2H2(g) + 5/2O2(g)--> 2CO2(g) + H2O(l) delta H =- 1299.6KJ/mole
-------------------------------------------------------------------------------
2C(s) +H2(g)---->C2H2(g) delta H=227.6 kJ.mole
which is the required equation
thus , the enthalpy change is 227.6 kJ
[Note:delta H is an extensive property and is dependent on amount.thus, the same operation is done on the delta H as what has been done with the reaction]
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