1. Tardigrade
  2. Question
  3. Chemistry
  4. For a reversible reaction X(g)+3Y(g)2Z(g); Δ H=-40 kJ, the standard entropies of X, Y and Z are 60,40 and 50 text JK-1mol-1 respectively. The temperature at which the above reaction attains equilibrium is about

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Q. For a reversible reaction $ X(g)+3Y(g)2Z(g); $ $ \Delta H=-40\,kJ, $ the standard entropies of X, Y and Z are 60,40 and $ 50\text{ }J{{K}^{-1}}mo{{l}^{-1}} $ respectively. The temperature at which the above reaction attains equilibrium is about

MGIMS WardhaMGIMS Wardha 2008

Solution:

$ {{X}_{(g)}}+3Y(g)2Z(g) $ $ \Delta {{s}^{o}}=2\Delta {{s}^{o}}(Z)-\{{{s}^{o}}(X)+3{{s}^{o}}(Y)\} $ $ =2\times 50-\{60+3\times 40\} $ $ =100-180 $ $ =-80\,J{{K}^{-1}}mo{{l}^{-1}} $ Given $ \Delta {{H}^{o}}=-40\,kJ $ $ =40,000\,J $ $ \Delta {{G}^{o}}=\Delta {{H}^{o}}-T\Delta {{S}^{o}} $ At equilibrium, $ \Delta {{G}^{o}}=0 $ $ \therefore $ $ \Delta {{H}^{o}}=T\Delta {{S}^{o}} $ or $ T=\frac{\Delta {{H}^{o}}}{\Delta {{s}^{o}}} $ $ =\frac{40,000}{80} $ $ =500K $