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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Sample Paper Topic Test - Chemical Equilibrium

  • question_answer
    \[{{H}_{2}}(g)+{{I}_{2}}(g)2HI(g)\] When 46g of \[{{I}_{2}}\] and 1g of \[{{H}_{2}}\] are heated at equilibrium at \[450{}^\circ C,\] the equilibrium mixture contained 1.9 g of of \[{{I}_{2}}\]. How many moles of  \[{{I}_{2}}\] and HI are present at equilibrium?

    A) 0.0075 & 0.147 moles

    B) 0.0050 & 0.147 moles

    C) 0.0075 & 0.347 moles

    D) 0.0052 & 0.347 moles

    Correct Answer: C

    Solution :

    [c] Moles of \[{{I}_{2}}\] taken \[=\frac{46}{254}=0.181\]
    \[\left( \because \text{no of moles}=\frac{\text{given mass}}{\text{molar mass}} \right)\]
    Moles of \[{{H}_{2}}\] taken \[=\frac{1}{2}=0.5\]
    Moles of \[{{I}_{2}}\] remaining \[=\frac{1.9}{254}=0.0075\]
    Moles of \[\]used \[=0.181-0.0075=0.1735\]
    Moles of \[{{\text{H}}_{\text{2}}}\] used \[=0.1735\]
    Moles of \[{{\text{H}}_{\text{2}}}\]remaining \[0.5-0.1735=0.3265\]
    Moles of HI formed \[=0.1735\times 2=0.347\]
    At equilibrium
    Moles of \[{{\text{I}}_{\text{2}}}=0.0075\]
    Moles of HI = 0.347 moles


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