Given the enthalpies of combustion of acetylene(C2H2), carbon and hydrogen, 2C 2

Question

Given the enthalpies of combustion of acetylene(C2H2), carbon and hydrogen,
2C2H2(g)+ 5O2(g) ?4CO2(g)+ 2H2O(l) ?Ho = -2600kJ

C(s) + O2(g) ? CO2(g) ?Ho = -394 kJ

2H2(g) + O2 ? 2 H2O(l) ?Ho = -572 kJ

Calculate the enthalpy of formation of acetylene. The reaction is shown below.
2C(s) + H2(g) ? C2H2(g)

Given the enthalpies of combustion of acetylene(C2H2), carbon and hydrogen, 2C2H2(g)+ 5O2(g) ?4CO2(g)+ 2H2O(l) ?Ho = -2600kJ C(s) + O2(g) ? CO2(g) ?Ho = -394 kJ 2H2(g) + O2 ? 2 H2O(l) ?Ho = -572 kJ Calculate the enthalpy of formation of acetylene. The reaction is shown below. 2C(s) + H2(g) ? C2H2(g)

Explanation / Answer

A-->E = A-->B-->C-->D-->E (the overall energy change remains the same, independently of the path chosen)


*btw, sorry about the horizontal lines, don't mind them, they're just to hold the arrows in place (you should focus on the upside-down triangle)


2C2H2(g) + 5O2(g) -----------> 4CO2(g) + 2H2O(l)
______^________-2600 kJ__^
_________________(h3)____/
_______________________/
_____(h1)______________/_(h2)
_____________________/
____________________/
______4C(s) + 2H2(g) + 5O2(g)


(h1) = formation of 2 moles of C2H2 from elements [we're looking for STANDARD formation enthalpy - meaning, the energy required to form 1 mole of C2H2 = (h1)/2]

(h2) = formation of 4 moles of CO2 and 2 moles of H2O = 4x(-394.0 kJ) + 2x(-572 kJ) = -2720 kJ

(h3) = -2600 kJ



now, (h1) = (h2) - (h3)

[we take the other path: we go from the elements, to the products of combustion and then to the reagents; since we reverse the arrow of (h3), we "reverse" the sign as well]

(h1) = -2720 kJ - (-2600 kJ)
(h1) = -2720 kJ + 2600 kJ
(h1) =-120 kJ

now, this is the energy required to form 2 moles of C2H2, so we divide by 2 to get the formation enthalpy of 1 mole

Standard formation enthalpy (C2H2) = -60 kJ/mol

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