Flag this question question 8 10 pts use the δh°f and δh°rxn information provided to calculate δh°f for if: δh°f (kj/mol) if7(g) + i2(g) → if5(g) + 2 if(g) δh°rxn = -89 kj if7(g) -941 if5(g) -840

Respuesta :

[tex] \Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \; \text{kJ} \cdot \text{mol}^{-1} [/tex]

Explanation

[tex] \text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g) [/tex]

  • [tex] \Delta H\textdegree{}_\text{rxn} = -89\; \text{kJ} \cdot \text{mol}^{-1} [/tex]
  • [tex] \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) = -941 \; \text{kJ} \cdot \text{mol}^{-1} [/tex] (from the question)
  • [tex] \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) = -840 \; \text{kJ} \cdot \text{mol}^{-1} [/tex] (from the question)
  • As an the most stable allotrope under standard conditions, [tex] \Delta H\textdegree{}_f (\text{I}_2) = 0\; \text{kJ} \cdot \text{mol}^{-1} [/tex]

By definition,

[tex] \Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products}) - \Delta H\textdegree{}_f (\text{all reactants}) [/tex]

[tex] \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + 2 \; \Delta H\textdegree{}_f (\text{IF} \; (g) ) - \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) - \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) \\ = \Delta H\textdegree{}_{\text{rxn}} [/tex]

[tex] [/tex] [tex] \begin{array}{ccc} \Delta H\textdegree{}_f (\text{IF} \; (g) )& = & 1/2\; ( \Delta H\textdegree{}_{\text{rxn}} - \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) + \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) )\\ & = & 1/2 \; (-89 - (-840) + (-941))}\\ & = & - 95 \; \text{kJ} \cdot \text{mol}^{-1} \end{array} [/tex]

Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the [tex] \Delta H\textdegree{} _f [/tex] given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

[tex] \Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \; \text{kJ} \cdot \text{mol}^{-1} [/tex] (Cox, Wagman, et al., 1984)

... yields [tex] \Delta H\textdegree{}_f (\text{IF} \; (g) ) \approx -64 \; \text{kJ}\cdot \text{mol}^{-1} [/tex], which deviates significantly from the experimental value of [tex] -94.76 \; \text{kJ}\cdot \text{mol}^{-1} [/tex] (Chase, 1998.) It is thus assumed that the [tex] \Delta H\textdegree{}_\text{rxn} [/tex] value provided requires a reaction with [tex] \text{I}_2 \; (s) [/tex] rather than [tex] \text{I}_2 \; (g) [/tex] as a reactant.

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