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NEET AIPMT SOLVED PAPER SCREENING 2009

  • question_answer
    From the following bond energies: H?H bond energy: 431.37 kJ \[\frac{dQ}{dt}\] C = O bond energy: 606.10 kJ\[\frac{dQ}{dt}=\frac{KL({{T}_{1}}-{{T}_{2}})}{A}\] C - C bond energy: 336.49 kJ\[\frac{dQ}{dt}=\frac{K({{T}_{1}}-{{T}_{2}})}{LA}\] C ? H bond energy: 410.50 kJ \[\frac{dQ}{dt}=KLA({{T}_{1}}-{{T}_{2}})\] Enthalpy for the reaction, \[\frac{dQ}{dt}=\frac{KA({{T}_{1}}-{{T}_{2}})}{L}\]will be

    A) 1523.6 kJ \[{{\varepsilon }_{1}}-({{i}_{1}}+{{i}_{2}})R-{{i}_{1}}{{r}_{1}}=0\]            

    B) - 243.6 kJ\[{{\varepsilon }_{2}}-{{i}_{2}}{{r}_{2}}-{{\varepsilon }_{1}}-{{i}_{1}}{{r}_{1}}=0\]            

    C) - 120.0 kJ\[-{{\varepsilon }_{2}}-({{i}_{1}}+{{i}_{2}})R+{{i}_{2}}{{r}_{2}}=0\]

    D) 553.0 kJ\[{{\varepsilon }_{1}}-({{i}_{1}}+{{i}_{2}})R+{{i}_{2}}{{r}_{2}}=0\]

    Correct Answer: C

    Solution :

                    Key Idea, Enthalpy of reaction \[\overrightarrow{E}=\hat{i}2xy+\hat{j}({{x}^{2}}+{{y}^{2}})+\hat{k}(3xz-{{y}^{2}})\] For reaction, \[\overrightarrow{E}=\hat{i}{{z}^{3}}+\hat{j}xyz+\hat{k}{{z}^{2}}\] \[\overrightarrow{E}=\hat{i}(2xy-{{z}^{3}})+\hat{j}x{{y}^{2}}+\hat{k}3{{z}^{2}}x\] \[2\times {{10}^{4}}J{{T}^{-1}}\] \[B=6\times {{10}^{4}}T\] \[60\Omega \] \[240\,\Omega \]


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