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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics2 Marks
Question
Use the following data to calculate the $\text{K}_{\text{sp}}$ value for Agl(s)$\text{H}_{2}\text{(g)} + \text{Cl}_{2} \text{(g)} \rightleftharpoons\text{2HCl(g)}$
$\Delta\text{H}^{\circ} = -92\text{kJ}, \ \Delta\text{S}^{\circ} = -95\text{JK}^{-1}$
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Answer

$\Delta \text{G}^{\circ} = \sum\Delta\text{G}^{\circ}_{\text{products}} - \sum\Delta\text{G}^{\circ}_{\text{reactants}}$$ = 77\text{kJ} - 52\text{kJ} - (-25.5\text{kJ})$
$= 50.5\text{kJ}$
$\Delta \text{G}^{\circ} = -2.303 \text{RT} \log \text{k}_{\text{sp}}$
$50.5 \text{kJ} = \frac{-2.303 \times 8.314\times 298\text{K} \log \text{K}_{\text{sp}}}{1000}$
$\log \text{K}_{\text{sp}} = \frac{-50.5\text{kJ}}{5.7058} $
$= -8.8506 + 1 - 1 = \bar{9}.1496$
$\text{K}_{\text{sp}} = \text{Antilog} \ \bar{9}.1496 = 1.411 \times 10^{-9}$

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