Answer:
ΔH = 53.28 kJ
Explanation:
Solving this kind of problems is quite straight forward. What is need is to manipulate the reactions by multiplying the coefficients and reversing if necessary the reactions so that when we add the reactions together at the end we will arrive to the desired equation we need to obtain the enthalpy.
The reaction we need to calculate the enthalpy is
3C(s) + 3H2(g) → C3H6(g) ΔH = ?
If we take the 2nd reaction and multiply it by three, the inverse of first one multiplied by 1/2 and three times the 3rd we will be get the desired equation and its enthalpy:
3 C(s) + 3 O2(g) → 3 CO₂(g) ΔH= 3 x ( -393.51 kJ)
3 CO₂(g) + 3 H₂O(l) → C₃H₆(g) + 9/2 O₂(g) ΔH= 1/2 x ( 4182.6 kJ)
3 H₂(g) + 3/2 O₂(g) → 3 H₂O(l) ΔH= 3 x ( -285.83 kJ)
3C(s) + 3H2(g) → C3H6(g)
Notice how the mole O2 cancel because they are in diferent sides of the equation. Also we changed the changed of the second since we inverted it.
ΔH = 3 x ( -393.51 kJ) + 1/2 x ( 4182.6 kJ) + 3 x ( -285.83 kJ)
ΔH = - 1180.53 + 2091.3 - 857.49 = 53.28 kJ
Answer:
[tex]\large \boxed{\text{53.3 kJ/mol}}[/tex]
Explanation:
We have three equations:
(I) 2C₃H₆(g) + 9O₂(g) → 6CO₂(g) + 6H₂O(l); ΔH = -4182.6 kJ/mol
(II) C(s) + O₂(g) → CO₂(g); ΔH = -393.51 kJ/mol
(III) H₂(g) + ½O₂(g) → H₂O(l); ΔH = -285.83 kJ/mol
From these, we must devise the target equation:
(IV) 3C(s) + 3H₂(g) → C₃H₆(g); ΔH = ?
The target equation has 3C on the left, so you triple Equation(II).
When you triple an equation, you triple its ΔH.
(V) 3C(s) + 3O₂(g) → 3CO₂(g); ΔH = -1180.53 kJ/mol
Equation V has 3CO₂ on the right, and that is not in the target equation.
You need an equation with 3CO₂ on the left, so you reverse Equation I and divide by 2.
When you reverse an equation, you reverse the sign of its ΔH.
When you divide an equation by 2, you divide its ΔH by 2.
(VI) 3CO₂(g) + 3H₂O(l) ⟶ C₃H₆(g) + ⁹/₂O₂(g); ΔH = 2091.3 kJ/mol
Equation VI has 3H₂O on the left, and that is not in the target equation.
You need an equation with 3H₂O on the right, so you multiply Equation III by 3.
(VII) 3H₂(g) + ³/₂O₂(g) → 3H₂O(l); ΔH = -857.49 kJ/mol
Now, you add equations V, VI, and VII, cancelling species that appear on opposite sides of the reaction arrows.
When you add equations, you add their ΔH values.
You get the target equation (III):
(V) 3C(s) + 3O₂(g) → 3CO₂(g); ΔH = -1180.53 kJ/mol
(VI) 3CO₂(g) + 3H₂O(l) ⟶ C₃H₆(g) + ⁹/₂O₂(g); ΔH = 2091.3 kJ/mol
(VII) 3H₂(g) + ³/₂O₂(g) ⟶ 3H₂O(l); ΔH = -857.49 kJ/mol
(III) 3C(s) + 3H₂(g) ⟶ C₃H₆(g) ΔH = 53.3 kJ/mol
[tex]\Delta H \text{ for the reaction is } \large \boxed{\textbf{53.3 kJ/mol}}[/tex]
