1. Tardigrade
  2. Question
  3. Chemistry
  4. For CaCO3(s)leftharpoons CaO(s)+CO2(g) at 927oC,Δ H =176 kJmol; then textE is:

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Q. For $ CaC{{O}_{3}}(s)\rightleftharpoons CaO(s)+C{{O}_{2}}(g) $ at $ 927{{}^{o}}C,\Delta H =176\,kJmol; $ then $ \text{E} $ is:

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Solution:

$ CaC{{O}_{3}}(g)\rightleftharpoons CaO(s)+C{{O}_{2}}(g) $ $ (\Delta {{n}_{g}}=1-0=1) $ We know that, $ \Delta H=\Delta E+\Delta {{n}_{g}}RT $
$ 176=\Delta E +1\times 8.31\times {{10}^{-3}}\times (273\,+927) $
$ \therefore $ $ \Delta E=176-8.31\times {{10}^{-31}}\times 1200 $
$ =176-9.97 $ $ =166.03\,\text{kJ} $