Consider the reactions
C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ∆H = −758 kJ mol−1
2C(s) + 2H2(g) → C2H4(g) ∆H = +52 kJ mol−1
H2(g) + O2(g) → H2O(g) ∆H = −242 kJ mol−1
The enthalpy of formation of carbon monoxide is
A −111 kJ mol−1
B −163 kJ mol−1
C −222 kJ mol−1
D -464 kJ mol−1
How would I go about this?
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Urgent as chem question about hess's law watch
- Thread Starter
- 11-02-2018 21:33
- 12-02-2018 08:58
Since you're given the enthalpy of formation for C2H4, you know the enthalpy of formation for the products (balance by adding O2)
Since you also know the enthalpy of combustion of hydrogen, you know the enthalpy of combustion of the products (save CO).
You can draw a cycle with the gaseous atoms as the shared common thing (O2, 2C, 2H2) and draw the corresponding arrows to the products and reactants from them (directed upwards to both).
Since you know the enthalpy change overall is - 758, you can construct the equation by following the arrows around (according to hess' law), and come up with (2CO + 2x-242)-(52) =-758, you can rearrange to find 2CO =-222, divide through by 2 to get CO =-111. I hope this makes sense