JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Sample Paper Topic Test - Chemical Kinetics

  • question_answer
    The inversion of cane sugar proceeds with half-life of 500 min at pH = 5 for any concentration of sugar. However, if pH = 6, the half-life changes to 50 min. The rate law expression for sugar inversion can be written as:

    A) \[r=K{{\left[ sugar \right]}^{2}}{{\left[ {{H}^{+}} \right]}^{0}}\]            

    B) \[r=K{{\left[ sugar \right]}^{1}}{{\left[ {{H}^{+}} \right]}^{0}}\]

    C) \[r=K{{\left[ sugar \right]}^{1}}{{\left[ {{H}^{+}} \right]}^{1}}\]

    D) \[r=K{{\left[ sugar \right]}^{0}}{{\left[ {{H}^{+}} \right]}^{0}}\]

    Correct Answer: C

    Solution :

    \[pH=5\] \[pH=6\]
    \[\left[ {{H}^{+}} \right]={{10}^{-5}}\] \[\left[ {{H}^{+}} \right]={{10}^{-6}}\]
    \[\frac{\left( {{t}_{1/2}} \right)}{\left( {{t}_{1/2}} \right)}={{\left( \frac{{{10}^{-5}}}{{{10}^{-6}}} \right)}^{1-n}}\] \[=\frac{1}{10}={{\left( 10 \right)}^{1-n}}\]
    \[{{10}^{-1}}={{\left( 10 \right)}^{1-n}}\,\] \[\Rightarrow \,1-n\,=-1\Rightarrow \,n=2\]


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