2. Solved Papers
  3. NEET

NEET AIPMT SOLVED PAPER MAINS 2010

  • question_answer
    The reaction,                                                                     \[2A(g)+B(g)3C(g)+D(g)\] is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

    A) \[[{{(0.75)}^{3}}(0.25)]\div [{{(1.00)}^{2}}(1.00)]\]

    B) \[[{{(0.75)}^{3}}(0.25)]\div [{{(0.50)}^{2}}(0.75)]\]

    C) \[[{{(0.75)}^{3}}(0.25)]\div [{{(0.50)}^{2}}(0.25)]\]

    D) \[[{{(0.75)}^{3}}(0.25)]\div [{{(0.75)}^{2}}(0.25)]\]

    Correct Answer: B

    Solution :

    \[\overset{{}}{\mathop{\begin{align}   & Initial \\  & Ar\,equil \\ \end{align}}}\,\underset{1-0.50}{\mathop{\underset{1}{\mathop{2A(g)}}\,}}\,\,\,\,+\underset{1-0.25}{\mathop{\underset{1}{\mathop{B(g)}}\,}}\,\,\,\,\,\underset{0.75}{\mathop{\underset{0}{\mathop{3C(g)}}\,}}\,\,\,+\,\,\,\,\underset{0.25}{\mathop{\underset{0}{\mathop{D(g)}}\,}}\,\]                \[K=\frac{{{(0.75)}^{3}}(0.25)}{{{(0.50)}^{2}}(0.75)}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner