Find the total enthalpy for rection
2C (g) + 3H2 (g) C2H6 (g)
Given the thermochemical equations.
2C(g) + 2 O2 (g) 2CO2 (g) ∆H = -786 kJ/mol
3H2 (g) + O2 (g) 3H2O (l) ∆H = -858 kJ/mol
2CO2 (g) + 3H2O (l) C2H6 (g) + O2 (g) ∆H = +1560 kJ/mol
eq - equation
(1st eq): 2C(g) + 2O2(g) → 2CO2(g), ∆H1 = −786 kJ/mol
(2nd eq): 3H2(g) + 1.5O2(g) → 3H2O(l), ∆H2 = −858 kJ/mol
(3d eq): 2CO2(g) + 3H2O(l) → C2H6(g) + 3.5O2(g), ∆H3 = +1560 kJ/mol
Target equation: 2C(g) + 3H2(g) → C2H6(g), ∆Hx = ???
Solution:
According to Hess's law, the heat of reaction depends upon initial and final conditions of reactants and does not depend of the intermediate path of the reaction.
1) Modify the three given equations to get the target equation:
(1st eq): do nothing. We need 2C(g) on the reactant side and that's what we have.
(2nd eq): do nothing. We need 3H2(g) on the reactant side and that's what we have.
(3d eq): do nothing. We need C2H6(g) on the product side and that's what we have.
2) Rewrite all three equations with the changes made (including changes in their enthalpy):
(1st eq): 2C(g) + 2O2(g) → 2CO2(g), ∆H1 = −786 kJ/mol
(2nd eq): 3H2(g) + 1.5O2(g) → 3H2O(l), ∆H2 = −858 kJ/mol
(3d eq): 2CO2(g) + 3H2O(l) → C2H6(g) + 3.5O2(g), ∆H3 = +1560 kJ/mol
3) Cancel out the common species on both sides:
3.5O2(g) ⇒ (sum of 1st eq and 2nd eq) & (3d eq)
3H2O(g) ⇒ (3d eq) & (2nd eq)
2CO2(g) ⇒ (3d eq) & (1st eq)
Thus, adding modified equations and canceling out the common species on both sides, we get:
2C(g) + 3H2(g) → C2H6(g)
Add the ΔH values of (1st), (2nd) and (3d) equations to get your answer:
∆Hx = ∆H1 + ∆H2 + ∆H3 = (−786 kJ/mol) + (−858 kJ/mol) + (+1560 kJ/mol) = −84 kJ/mol
Therefore, the enthalpy of reaction is −84 kJ/mol
Answer: The total enthalpy for rection is −84 kJ/mol
Comments
Leave a comment