Hey guys! I'm really lost on this equation. Any help?
H2(g) yIelds 2H(g)
Delta H = 436 kJ/mol
F2(g) yields 2F(g)
DeltaH = 158 kJ/mol
H(g) + F(g) yields HF(g) DeltaH = -568 kJ/mol
Calculate deltaH for:
H2(g) + F2(g) yields 2 HF(g)
1 H2(g) yIelds 2H(g)
Delta H = 436 kJ/mol
2 F2(g) yields 2F(g)
DeltaH = 158 kJ/mol
3 H(g) + F(g) yields HF(g) DeltaH = -568 kJ/mol
Calculate deltaH for:
H2(g) + F2(g) yields 2 HF(g)
Bold face I have numbered the equations.
Add eqn 1 to eqn 2 and add twice the equation 3. That gives the equation you want. So dH rxn will be dH1 + dH2 + 2*dH3 = ?
To calculate ΔH for the given reaction, which is H2(g) + F2(g) → 2 HF(g), we can use Hess's law. Hess's law states that the enthalpy change of a reaction is the same whether the reaction occurs in one step or multiple steps.
First, let's break down the given reaction into a series of reactions that we already have enthalpy data for:
1. H2(g) → 2H(g) ΔH = 436 kJ/mol
2. F2(g) → 2F(g) ΔH = 158 kJ/mol
3. H(g) + F(g) → HF(g) ΔH = -568 kJ/mol
By algebraic manipulation, we can rearrange these reactions to match the desired reaction:
1. Reverse the first equation: 2H(g) → H2(g) ΔH = -436 kJ/mol
2. Reverse the second equation: 2F(g) → F2(g) ΔH = -158 kJ/mol
3. Reverse and multiply the third equation by 2: 2HF(g) → H(g) + F(g) ΔH = 2 * (-568 kJ/mol) = -1136 kJ/mol
Now, let's add up these reactions to obtain the desired reaction. Keep in mind that we can multiply the enthalpy values by any coefficient:
Desired reaction: H2(g) + F2(g) → 2 HF(g)
Adding the reactions:
-436 kJ/mol (2H(g)) + (-158 kJ/mol) (2F(g)) + (-1136 kJ/mol) (2HF(g))
Simplifying the equation, we get:
ΔH = -872 kJ/mol - 316 kJ/mol - 2272 kJ/mol = -3460 kJ/mol
Therefore, the ΔH for the reaction H2(g) + F2(g) → 2 HF(g) is -3460 kJ/mol.