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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Calculate the value of equilibrium constant for a reaction at 400K \[\] $2NOCl(g)2NO(g) + Cl(g)$ \[\] $ \Delta S$ and $ \Delta H$ are 122 J/Kmol and 77.2 kJ/mol respectively.

$\begin {array} {1 1} (A)\;1.96 \times 10^{-4} & \quad (B)\;19.6 \times 10^{-4} \\ (C)\;19.6 \times 10^{-2} & \quad (D)\;1.96 \times 10^{-2} \end {array}$

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$ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} = 77200\: – 400(122) = 28400\: J/mol$
$ \Delta G^{\circ}= -2.303\: RT \: log \: K $
$ \Rightarrow log\: K = \large\frac{- 28400}{(2.303 \times 8.314 \times 400)}$$ = 4.292$
$ \Rightarrow K = 1.96 \times10^{-4}$
Ans : (A)
answered Mar 17, 2014 by thanvigandhi_1
 

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