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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    The following reaction is performed at \[298K.2NO(g)+{{O}_{2}}2N{{O}_{2}}(g)\] The standard free energy of formation of \[NO(g)\] is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of \[N{{O}_{2}}(g)\] at 298K? \[\left( {{K}_{p}}=1.6\times {{10}^{12}} \right)\] [JEE Main Solved Paper-2015 ]

    A) \[86600-\frac{\ln \left( 1.6\times {{10}^{12}} \right)}{R\left( 298 \right)}\]

    B) \[0.5\left[ 2\times 86,600-R\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right) \right]\]

    C) \[R\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right)-86600\]

    D) \[86600+R\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right)\]

    Correct Answer: B

    Solution :

    \[\Delta {{G}_{R}}=-RT\ln Kp\] \[\Delta {{G}_{R}}=2{{\Delta }_{f}}{{G}_{N{{O}_{2}}}}-2\Delta {{G}^{0}}_{NO}\] \[-RT\ln 1.6\times {{10}^{12}}=2\left( {{\Delta }_{f}}{{G}_{N{{O}_{2}}}}-86,600 \right)\] \[86,600\frac{RT\ln \left( 1.6\times {{10}^{12}} \right)}{2}={{\Delta }_{f}}{{G}_{N{{O}_{2}}}}\] \[{{\Delta }_{f}}{{G}_{N{{O}_{2}}}}=0.5\left[ 2\times 86600-R\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right) \right]\]


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