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JEE Main & Advanced Sample Paper JEE Main Sample Paper-37

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

    A)  \[7.14\]                        

    B)  \[14.28\]

    C)  \[28.57\]                      

    D)  \[33.33\]                

    Correct Answer: B

    Solution :

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


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