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Manipal Medical Manipal Medical Solved Paper-2014

  • question_answer
    The equilibrium constant for the isomerisation of butane at25° C is \[{{K}_{c}}\]= 8 Butane \[\to \] Isobutane If 5.8 g butane is introduced into a 12.5 L vessel at 25° C. What mass of isobutane will be present at equilibrium?

    A) 5.15g                                        

    B) 2.56g

    C) 7.49g                                        

    D)  5.80g

    Correct Answer: A

    Solution :

    Initial mole of butane \[=\frac{5.8}{58}=0.1\] \[\underset{0.1-x}{\mathop{Bu\tan e}}\,\rightleftharpoons \underset{x}{\mathop{Isobu\tan e}}\,\] \[{{K}_{C}}=8=\frac{x}{0.1-x}\] or \[x=0.088\] m(isobutene) \[=0.088\times 58=5.15g\]


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