2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    At 675K, \[{{\text{H}}_{\text{2}}}\] (g) and \[\text{C}{{\text{O}}_{\text{2}}}\] (g) react to form CO(g) and \[{{\text{H}}_{\text{2}}}\text{O}\](g), \[{{\text{K}}_{\text{p}}}\] for the reaction is 0.16. If a mixture of 0.25 mole of \[{{\text{H}}_{\text{2}}}\] (g) and 0.25 mol of \[\text{C}{{\text{O}}_{\text{2}}}\] is heated at 675K, mole % of CO (g) in equilibrium mixture is

    A)  7.14                      

    B)  14.28

    C)  28.57                   

    D)  33.33

    Correct Answer: B

    Solution :

     \[\underset{0.25-x}{\mathop{{{H}_{2}}(g)+}}\,\underset{0.25-x}{\mathop{C{{O}_{2}}(g)}}\,CO\underset{x}{\mathop{(g)}}\,+{{H}_{2}}O\underset{x}{\mathop{(g)}}\,\] At \[\text{e}{{\text{q}}^{\text{m}}}\] \[{{K}_{p}}=0.16=\frac{{{x}^{2}}}{{{(0.25-x)}^{2}}}\Rightarrow 0.4=\frac{x}{0.25-x}\] \[\Rightarrow 0.1-0.4x=x\] \[x=0.0714\] Mole % of CO \[(g)=\frac{0.0714}{0.50}\times 100=14.28\]


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