Use the thermochemical equations:
1/2 H2(g)+1/2 I2(g) ...............> HI(g) delta H = -1.1 k cal
1/2 H2(g).......> H2 (g) delta H = -104.2 k cal
1/2 I2 (g)..............> I (g) delta H = +18.1 k cal
To present for the formation of hydrogen iodide gas from gaseous atomic hydrogen and gaseous atomic iodine:
H(g) +I(g) HI(g) delta H=?
Use Hess' Law.
Use equation 1 as is.
Add to the reverse of equation 2 (change the sign of delta H here) and add to the reverse of equation 3 (change the sign of delta H for 3 also).
You should get H + I ==> HI, then add all of the delta Hs together to arrive at a delta H for the reaction.
To determine the enthalpy change (delta H) for the formation of hydrogen iodide (HI) gas from gaseous atomic hydrogen (H) and gaseous atomic iodine (I), we can use the thermochemical equations given.
The given equations are:
1/2 H2(g) + 1/2 I2(g) → HI(g) ΔH = -1.1 kcal
1/2 H2(g) → H(g) ΔH = -104.2 kcal
1/2 I2(g) → I(g) ΔH = +18.1 kcal
To form HI, we need H and I, so we will add the two equations involving H and I to obtain the equation:
(1/2 H2(g) → H(g)) + (1/2 I2(g) → I(g)) → HI(g)
To combine the equations, we add the individual enthalpy changes (delta H) for each step. The coefficients of the equations need to be adjusted to ensure that the reactants and products match:
1/2 H2(g) + 1/2 I2(g) → HI(g) ΔH = -1.1 kcal
-2 × (1/2 H2(g) → H(g)) ΔH = 2 × (-104.2 kcal)
-2 × (1/2 I2(g) → I(g)) ΔH = 2 × (+18.1 kcal)
The adjusted equation becomes:
- H2(g) - I2(g) → 2HI(g) ΔH = -1.1 kcal + (-2 × -104.2 kcal) + (2 × 18.1 kcal)
Now, let's calculate the delta H for the formation of HI:
ΔH = -1.1 kcal + (-2 × -104.2 kcal) + (2 × 18.1 kcal)
= -1.1 kcal + 208.4 kcal + 36.2 kcal
= 243.5 kcal
Therefore, the delta H for the formation of hydrogen iodide gas from gaseous atomic hydrogen and gaseous atomic iodine is 243.5 kcal.