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JEE Main & Advanced Sample Paper JEE Main Sample Paper-11

  • question_answer
    The \[{{\Delta }_{f}}{{H}^{o}}({{N}_{2}}{{O}_{5}},g)\]in kJ/mol on the basis of the following data is - \[\text{2NO(g)+}{{\text{O}}_{\text{2}}}\text{(g)}\to \text{2N}{{\text{O}}_{\text{2}}}\text{(g);}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{o}}}\text{=-114kJ/mol}\] \[\text{4N}{{\text{O}}_{\text{2}}}\text{(g)+}{{\text{O}}_{\text{2}}}\text{(g)}\to \text{2}{{\text{N}}_{\text{2}}}{{\text{O}}_{5}}\text{(g);}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{o}}}\text{=-102}\text{.6kJ/mol}\] \[{{\text{ }\!\!\Delta\!\!\text{ }}_{f}}{{\text{H}}^{\text{o}}}\text{(NO,g)=90}\text{.2}\,\text{kJ/mol}\]

    A)  15.1                      

    B)  30.2

    C)  -36.2                    

    D)  None of these

    Correct Answer: A

    Solution :

     \[\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}NO(g);{{\Delta }_{f}}{{H}^{o}}=90.2\] \[{{N}_{2}}(g)+{{O}_{2}}\to 2NO(g);{{\Delta }_{f}}H{}^\circ =90.2\times 2\]          ? (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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