Given the enthalpies of combustion of propane (C3H8), carbon and hydrogen, C 3 H

Question

Given the enthalpies of combustion of propane (C3H8), carbon and hydrogen,
C3H8(g)+ 5O2(g) ? 3CO2(g)+ 4H2O(l) ?Ho = -2219.9kJ

C(s)  +  O2(g)   ?   CO2(g)         ?Ho = -393.5 kJ

2H2(g)  +  O2   ? 2 H2O(l)       ?Ho = -571.6 kJ



Calculate the enthalpy of formation of propane.  The reaction is shown below.


3C(s)  +   4H2(g)   ?   C3H8(g) Question options: -103.8 kJ 467.8 kJ
3185 kJ Given the enthalpies of combustion of propane (C3H8), carbon and hydrogen,
C3H8(g)+ 5O2(g) ? 3CO2(g)+ 4H2O(l) ?Ho = -2219.9kJ

C(s)  +  O2(g)   ?   CO2(g)         ?Ho = -393.5 kJ

2H2(g)  +  O2   ? 2 H2O(l)       ?Ho = -571.6 kJ



Calculate the enthalpy of formation of propane.  The reaction is shown below.


3C(s)  +   4H2(g)   ?   C3H8(g) Given the enthalpies of combustion of propane (C3H8), carbon and hydrogen,
C3H8(g)+ 5O2(g) ? 3CO2(g)+ 4H2O(l) ?Ho = -2219.9kJ

C(s)  +  O2(g)   ?   CO2(g)         ?Ho = -393.5 kJ

2H2(g)  +  O2   ? 2 H2O(l)       ?Ho = -571.6 kJ



Calculate the enthalpy of formation of propane.  The reaction is shown below.


3C(s)  +   4H2(g)   ?   C3H8(g) -103.8 kJ 467.8 kJ
3185 kJ -103.8 kJ 467.8 kJ 3185 kJ -103.8 kJ 467.8 kJ
3185 kJ

Explanation / Answer

3CO2(g)+ 4H2O(l) ? C3H8(g)+ 5O2(g) ?Ho = +2219.9kJ

3C(s) + 3O2(g) ? 3CO2(g) ?Ho = -393.5*3 kJ

4H2(g) + 2O2 ? 4 H2O(l) ?Ho = -571.6*2 kJ

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3C(s) + 4H2(g) ? C3H8(g) ?Ho=2219.9-(393.5*3)-(571.6*2)= -103.8KJ

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