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JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    The reaction \[2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g)\]follows first order kinetics. The pressure of a vessel containing only \[{{N}_{2}}{{O}_{5}}\] was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. will be (Assume temperature remains constant) : JEE Main Online Paper (Held On 10 April 2015)

    A)  150 mm He

    B)  116.05 mm Hg

    C)   106.25 mm Hg

    D)  125 mm Hg

    Correct Answer: C

    Solution :

                     First order reaction rate \[=K\left[ {{N}_{2}}{{O}_{5}} \right]\] \[2{{N}_{2}}{{O}_{5(g)}}\xrightarrow[{}]{{}}4N{{O}_{2(g)}}+{{O}_{2(g)}}\] t = 0 pressure 50                               0              0 t = 30 t pressure 50- 2p                  4p           p Total pressure \[50-2p+4p+p=50+3p=87.5mmHg\] \[\therefore \] \[P=12.5mmHg\] \[\therefore \]\[{{P}_{0}}=50\And {{P}_{t}}=25\]for\[{{N}_{2}}{{O}_{5}}\]reactant \[\therefore \]\[K=\frac{2.303}{30\min }\times \log \left( \frac{50}{25} \right)=\frac{2.303}{60\min }\times \log \left( \frac{50}{x} \right)\]On solving \[x=12.5mmHg=50-2p\] \[\therefore \]\[P=18.75mmHg\] \[\therefore \]Total pressure\[=50+3p=106.25mmHg\]


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