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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2016 · Chemistry · Thermodynamics (C)

Given F(g)+ ^ - arrow F^ - (g) (Electron gain enthalpy) Li^ +...

Given Given F(g)+ ^ - arrow F^ - (g) (Electron gain enthalpy) Li^ + (g)+F^ - (g) arrow LiF(s) -1047Li(s)+(1)/(2) F_ 2 ( g)
F(g)+\mathrme−→F−(g)\mathrm{F}(\mathrm{g})+\mathrme^{-} \rightarrow \mathrm{F}^{-}(\mathrm{g}) \quad (Electron gain enthalpy) Li+(g)+F−(g)→LiF(s)−1047Li(s)+12F2(g)→LiF(s)−617\mathrm{Li}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) \rightarrow \mathrm{LiF}(\mathrm{s}) \quad-1047\mathrm{Li}(\mathrm{s})+\frac{1}{2} \mathrm{ F}_{2}(\mathrm{ g}) \rightarrow \mathrm{LiF}(\mathrm{s}) \quad-617 Based on data provided, the value of electron gain enthalpy of fluorine would be:

Options

  1. A−300kJmol−1-300 \mathrm{ kJ} \mathrm{ mol}^{-1}
  2. B−350kJmol−1-350 \mathrm{ kJ} \mathrm{ mol}^{-1}
  3. C−328kJmol−1-328 \mathrm{ kJ} \mathrm{ mol}^{-1}
  4. D−228kJmol−1-228 \mathrm{ kJ} \mathrm{ mol}^{-1}

Correct answer

C. −328kJmol−1-328 \mathrm{ kJ} \mathrm{ mol}^{-1}

Step-by-step solution

Applying Hess's Law ΔfH∘=Δsub mathrmH+12ΔdissH+L.E. +E.A+Δlattice H−617=161+520+77+EA.+(−1047)E.A.=−617+289=−328kJmol−1\begin{array}{c} \Delta_{\mathrm{f}} \mathrm{H}^{{}^{\circ}}=\Delta_{\mathrm{sub}}\ mathrm{H}+\frac{1}{2} \Delta_{\mathrm{diss}} \mathrm{H}+\mathrm{L} \text{.E. }+\mathrm{E}. \mathrm{A}+\Delta_{\text{lattice }} \mathrm{H}\\ -617=161+520+77+\mathrm{E} \mathrm{A}.+(-1047) \\ \mathrm{E}. \mathrm{A}.=-617+289=-328 \mathrm{ kJ} \mathrm{ mol}^{-1} \end{array}\thereforeelectronaffinityoffluorineelectron affinity of fluorine=-328 \mathrm{ kJ} $$\mathrm{ mol}^{-1}
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