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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    At\[\text{518  }\!\!{}^\circ\!\!\text{ C}\], die rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr \[{{\text{s}}^{-1}}\] when 5% had reacted and, 0.5 Torr \[{{\text{s}}^{-1}}\] when 33% had reacted. The order of me reaction is:   [JEE Main Online 08-04-2018]

    A)  1                                

    B)  0

    C)  2                                

    D)  3

    Correct Answer: C

    Solution :

                      rate\[\text{=-}\frac{\text{dP}}{\text{dt}}\text{=k}\left( {{\text{P}}_{\text{C}{{\text{H}}_{\text{3}}}\text{CHO}}} \right)\] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{1}{0.5}=\frac{{{\left( \frac{363\times 95}{100} \right)}^{x}}}{{{\left( 363\times \frac{67}{100} \right)}^{x}}}\] \[2={{(1.41)}^{x}}\]         \[2={{(\sqrt{2})}^{x}}\] \[\]


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