2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-20

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

    A)  \[15.1\]                        

    B)  \[30.2\]

    C)  \[-36.2\]                       

    D)  None of these

    Correct Answer: A

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner