1. Tardigrade
  2. Question
  3. Chemistry
  4. Using the following thermochemical equations (i) S(rh)+3/2 O2(g) xrightarrow[]SO3(g) Δ H=-2x kJ mol-1 (ii) SO2(g)+1/2 O2(g) xrightarrowSO3(g) Δ H=-y kJ mol-1 . Find out the heat of formation of SO2(g) in kJ mol-1 .

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Q. Using the following thermochemical equations
(i) $ S_{(rh)}+3/2\,{{O}_{2}}_{(g)}\xrightarrow[{}]{{}}S{{O}_{3}}_{(g)} $ $ \Delta H=-2x\,kJ\,mo{{l}^{-1}} $
(ii) $ S{{O}_{2}}_{(g)}+1/2\,{{O}_{2}}_{(g)}\xrightarrow{{}}S{{O}_{3}}_{(g)} $ $ \Delta H=-y\,kJ\,mo{{l}^{-1}} $.
Find out the heat of formation of $ S{{O}_{2}}(g) $ in $kJ$ $ mo{{l}^{-1}} $ .

KEAMKEAM 2008Thermodynamics

Solution:

Given $ S(rh)+\frac{3}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}S{{O}_{3}}(g); $
$ \Delta H=-2x\,kJ\,mo{{l}^{-1}} $ ...(i)
$ S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}S{{O}_{3}}(g); $
$ \Delta H=-y\,kJ\,mo{{l}^{-1}} $ ...(ii)
$ S(s)+{{O}_{2}}(g)\xrightarrow{{}}S{{O}_{2}}(g); $
$ \Delta H=? $ Subtract Eq (ii) from Eq (i) $a S(rh)+\frac{3}{2}{{O}_{2}}(g)\xrightarrow{{}}S{{O}_{3}}(g); $ $ \Delta H=-2x\text{ }kJ\text{ }mo{{l}^{-1}} $ $ S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}S{{O}_{3}}(g); $ $ \Delta H=-y\,kJ\,mo{{l}^{-1}} $ $ \underline{-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,} $ $ S(rh)+\left( \frac{3}{2}-\frac{1}{2} \right){{O}_{2}}(g)\xrightarrow{{}}S{{O}_{2}}(g); $ $ \Delta H=(-2x+y) $ $ kJ\,mo{{l}^{-1}} $ $ S(rh)+{{O}_{2}}(g)\xrightarrow[{}]{{}}S{{O}_{2}}(g); $ $ \Delta H=(y-2x)kJ\,mo{{l}^{-1}} $