2. Solved Papers
  3. JAMIA MILLIA ISLAMIA

JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2009

  • question_answer
        A mixture of 0.3 mol of H2 and 0. 3 mole of 12 is allowed to react in a 10 L evacuated flask at\[500{}^\circ C\]. The reaction is\[{{H}_{2}}+{{I}_{2}}2HI\].The K is found to be 64. The amount of unreacted 12 at equilibrium is

    A)  0.15 mol                             

    B)  0.06 mol

    C)  0.03 mol                             

    D)  0.2 mol

    Correct Answer: B

    Solution :

                    Let\[x\]mol each of\[{{H}_{2}}\]and\[{{I}_{2}}\]react to give\[2x\]mole of\[HI\]at equilibrium. Then at equilibrium \[[{{H}_{2}}]=[{{I}_{2}}]=(0.3-x)/10\] \[\frac{{{[2n/10]}^{2}}}{\left[ \frac{0.3-x}{10} \right]\left[ \frac{0.3-x}{10} \right]}=\frac{4{{x}^{2}}}{{{(0.3-x)}^{2}}}\] Or           \[8=\frac{2x}{0.3-x}=x=0.24\] Mole of\[{{I}_{2}}\]left unreacted\[=0.3-0.24=0.06\]


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