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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    Given that \[C+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}};\Delta {{H }^{o}}=-x\,kJax\] \[2CO+{{O}_{2}}\xrightarrow{{}}2C{{O}_{2}};\Delta {{H}^{o}}=-y\,kJ\] The heat of formation of carbon monoxide will be

    A)  \[\frac{y-2x}{2}\]

    B)  \[y+2x\]

    C)  \[2x-y\]

    D)  \[\frac{2x-y}{2}\]

    Correct Answer: A

    Solution :

     Given, \[C+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}};\Delta {{H }^{o}}=-x\,kJ\] ?(i) \[2CO+{{O}_{2}}\xrightarrow{{}}2C{{O}_{2}};{{H}^{o}}=-Y\,kJ\] ?(ii) Required equation is \[C+\frac{1}{2}{{O}_{2}}\xrightarrow{{}}CO;\Delta H _{1}^{o}=?\] On reversing eq. (ii), we get \[2C{{O}_{2}}\xrightarrow{{}}2CO+{{O}_{2}};\Delta H_{2}^{o}=+Y\,kJ\]     ?(iii) Dividing eq. (iii) by 2 gives \[C{{O}_{2}}\xrightarrow{{}}CO+\frac{1}{2}{{O}_{2}};\Delta H_{3}^{o}=+\frac{Y}{2}\,kJ\] ?(iv) On adding eq. (i) and (iv), we get (required equation) \[C+\frac{1}{2}{{O}_{2}}\xrightarrow{{}}CO;\Delta H_{1}^{o}=\Delta {{H}^{o}}+\Delta H_{3}^{o}\] \[=\left( -x\frac{y}{2} \right)\,kJ\] \[=\frac{y-2x}{2}\,kJ\]


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