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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    Given \[{{C}_{(grahite)}}+{{O}_{2}}(g)\to C{{O}_{2}}(g);\] \[{{\Delta }_{r}}{{H}^{o}}=-393.5\,kJ\,mo{{l}^{-1}}\] \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(l);\] \[{{\Delta }_{r}}{{H}^{o}}=-285.8\,kJ\,mo{{l}^{-1}}\] \[C{{O}_{2}}(g)+2{{H}_{2}}O(l)\to C{{H}_{4}}(g)+2{{O}_{2}}(g);\] \[{{\Delta }_{r}}{{H}^{o}}=+890.3\,kJ\,mo{{l}^{-1}}\] Based on the above thermochemical equations, the value of \[{{\Delta }_{r}}{{H}^{o}}\]at 298 K for the reaction\[{{C}_{(grahite)}}+2{{H}_{2}}(g)\to C{{H}_{4}}(g)\]will be:-     JEE Main Solved Paper-2017

    A)  \[+\,74.8\,kJ\,mo{{l}^{-1}}\]     

    B) \[+\,144.0\,\,kJ\,mo{{l}^{-1}}\]

    C)  \[-74.8\,\,kJ\,mo{{l}^{-1}}\]      

    D)  \[-144.0\,kJ\,mo{{l}^{-1}}\]

    Correct Answer: C

    Solution :

    \[C{{O}_{2}}(g)+2{{H}_{2}}O(\ell )\to C{{H}_{4}}(g)+2{{O}_{2}}(g);{{\Delta }_{r}}{{H}^{o}}=890.3\]                    \[{{\Delta }_{f}}{{H}^{o}}-393.5-285.8\,\,\,\,?\,\,\,\,\,\,0\] \[{{\Delta }_{r}}{{H}^{o}}={{\sum{({{\Delta }_{f}}{{H}^{o}})}}_{\text{product}}}-{{\sum{({{\Delta }_{f}}{{H}^{o}})}}_{\text{reactan}\,\text{ts}}}\]\[890.3=\left[ 1\times {{({{\Delta }_{f}}{{H}^{o}})}_{C{{H}_{4}}}}+2\times 0 \right]-[1\times (-393.5)+2(-285.8)]\]\[{{({{\Delta }_{f}}{{H}^{o}})}_{C{{H}_{4}}}}=890.3-965.1=-74.8\,kJ/mol\]


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