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JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    A closed vessel initially contains a mixture of two gases \[{{A}_{2}}B\]and an in \[{{A}_{2}}B\]gas. During the thermal decomposition of gas A, B as per given \[{{A}_{2}}B(g)\to 2A(g)+B(g)\] the pressure changed from an initial value of 2 atm to 5.2 atm at the end of reaction. The rate constant (in min) if total pressure was measured as 4.8 atm after 10 min is-

    A) \[0.3\,\log 2\]                   

    B) \[3\ln \frac{1.6}{1.6-1.4}\]

    C) \[\frac{\ln 2}{200}\]                                       

    D) \[\frac{3\ln 2}{10}\]

    Correct Answer: D

    Solution :

    Let pressure due to to \[{{A}_{2}}B(g)\]be ?x? atm \[\therefore \]pressure due to inert gas = (2 - x) atm
    \[{{A}_{2}}B(g)\]\[\to \] \[2A(g)\]\[+\] \[B(g)\]
    \[t=0\] \[x\] \[0\] \[0\]
    \[t={{t}_{end}}\] \[0\] \[2x\] \[x\]
    \[\therefore \]Total pressure\[=2x+x(2-x)=5.2\] \[\therefore \]\[x=1.6\,atm\] After 10 mins: -
    \[{{A}_{2}}B(g)\]\[\to \] \[2A(g)\]\[+\] \[B(g)\]
    \[t=10\,\min \] \[1.6-y\] \[2y\] y
    Total pressure \[=(1.6-y)+2y+y+(2-1.6)=4.8\] \[\therefore \]\[2y+2=4.8\] \[\therefore \]\[y=1.4\,atm\] \[\therefore \]\[K=\frac{1}{10}\ln \frac{x}{x-y}=\frac{1}{10}\ln \left( \frac{1.6}{0.2} \right)=\frac{3\ln 2}{10}\]


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