Question

if the standard enthalpy change for the reaction below is 157 kj/mol. calculate the standard enthalpy of formation of atomic fluorine (f). round your answer to 3 significant digits.
f₂(g) → f(g) + f(g)
note: reference the thermodynamic properties of pure substances table for additional information.

Explanation:

Step1: Relate reaction enthalpy to formation enthalpies

The standard enthalpy change of a reaction ($\Delta H^\circ$) is the sum of the standard enthalpies of formation of the products minus the sum of those of the reactants. For the reaction $\text{F}_2(g)
ightarrow 2\text{F}(g)$, this is:
$$\Delta H^\circ = 2\Delta H_f^\circ(\text{F}(g)) - \Delta H_f^\circ(\text{F}_2(g))$$

Step2: Use $\Delta H_f^\circ$ of elements in standard state

The standard enthalpy of formation of an element in its standard state is 0. $\text{F}_2(g)$ is the standard state of fluorine, so $\Delta H_f^\circ(\text{F}_2(g)) = 0\ \frac{\text{kJ}}{\text{mol}}$.

Step3: Solve for $\Delta H_f^\circ(\text{F}(g))$

Given $\Delta H^\circ = 157\ \frac{\text{kJ}}{\text{mol}}$, substitute into the equation:
$$157 = 2\Delta H_f^\circ(\text{F}(g)) - 0$$
$$\Delta H_f^\circ(\text{F}(g)) = \frac{157}{2} = 78.5$$

Step4: Round to 3 significant digits

78.5 has exactly 3 significant digits.

Answer:

78.5