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NEET Sample Paper NEET Sample Test Paper-14

  • question_answer
    The free energy formation of NO is \[78\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K? \[\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}^{2}}(g)\Leftrightarrow NO(g)\]

    A) \[8.4\times {{10}^{-5}}\]                

    B) \[7.1\times {{10}^{-9}}\]

    C) \[4.2\times {{10}^{-10}}\]              

    D) \[1.7\times {{10}^{-19}}\]  

    Correct Answer: A

    Solution :

    \[{{K}_{eq}}=\text{antilog}\] \[\Delta {{G}^{o}}=78\,kJ/mol=78000\,J/mol.\] Putting these values in equation 1 we get, \[{{K}_{eq}}=\text{antilog}\] \[{{K}_{eq}}=\text{anitilog}\,\text{(-4}\text{.0737)}\]or \[{{K}_{eq}}=8.4\times {{10}^{-5}}.\] Hence, the correct option is [a].


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