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JEE Main & Advanced Sample Paper JEE Main Sample Paper-8

  • question_answer
    The in \[{{\Delta }_{f}}H{}^\circ ({{N}_{2}}{{O}_{5}},g)\] kJ/mol on the basis of the following data is - \[2NO(g)+{{O}_{2}}(g)\to 2N{{O}_{2}}(g);\,\,{{\Delta }_{r}}H{}^\circ =-114\,kJ/mol\]\[4N{{O}_{2}}(g)+{{O}_{2}}(g)\to 2{{N}_{2}}{{O}_{5}}(g);\,\,{{\Delta }_{r}}H{}^\circ =-102.6\,kJ/mol\]\[{{\Delta }_{f}}H{}^\circ (NO,g)=90.2kJ/mol\]

    A)   15.1                    

    B)  30.2

    C)  -36.2                    

    D)  None of these

    Correct Answer: A

    Solution :

     \[\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}NO(g);{{\Delta }_{f}}{{H}^{o}}=90.2\] \[{{N}_{2}}(g)+{{O}_{2}}(g)\xrightarrow[{}]{{}}2NO(g);{{\Delta }_{f}}{{H}^{o}}=90.2\times 2\]?(1) \[2NO(g)+{{O}_{2}}(g)\xrightarrow[{}]{{}}2N{{O}_{2}}(g);{{\Delta }_{f}}{{H}^{o}}=-114\]?..(2) \[2N{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}{{N}_{2}}{{O}_{5}}(g);\] \[\Delta f{{H}^{o}}=\frac{-102.6}{2}=-51.3\]                                         ?(3) \[Eq.(1)+(2)+(3)\] \[{{N}_{2}}(g)+\frac{5}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}{{N}_{2}}{{O}_{5}}(g)\] \[{{\Delta }_{f}}{{H}^{o}}({{N}_{2}}{{O}_{5}},g)=15.1\text{kJ/mol}\]


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