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Kryger [21]
3 years ago
10

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
Chemistry
1 answer:
GarryVolchara [31]3 years ago
3 0

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

Explanation

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

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

By definition,

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

\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}}

\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}

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 \Delta H\textdegree{} _f 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

\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} (Cox, Wagman, et al., 1984)

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

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