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J & K CET Engineering J and K - CET Engineering Solved Paper-2011

  • question_answer
    The standard enthalpy of formation of \[{{C}_{2}}{{H}_{4}}(g),C{{O}_{2}}(g)\] and\[{{H}_{2}}O(l)\]are\[~52,-394\] and \[-284\,kJ\,mo{{l}^{-1}}\]respectively. Then the amount of heat evolved by burning 7 g of \[{{C}_{2}}{{H}_{4}}(g)\]is

    A)  1412 kJ        

    B)  9884 kJ

    C)  353 kJ         

    D)  706 kJ

    Correct Answer: C

    Solution :

     \[{{C}_{2}}{{H}_{4}}(g)+3{{O}_{2}}(g)\xrightarrow{{}}2C{{O}_{2}}(g)+2{{H}_{2}}O(l);\] \[\Delta H=?\] \[\Delta H=\sum{\Delta {{\Eta }_{f}}}\]product \[-\sum{\Delta {{\Eta }_{f}}}\]reactant \[=[2(-394)+2(-286)]=[(52+0)]\] \[=[2(-394)+2(-286)]-[(52+0)]\] \[=[-788-572]-[52]\] \[=-1412\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\] \[\because \]Amount of heat evolved by burning 28 g of\[{{C}_{2}}{{H}_{4}}=1412\,kJ\] \[\therefore \]Amount of heat evolved by burning 7 g of \[{{C}_{2}}{{H}_{4}}=\frac{1412}{28}\times 7=353\,kJ\]


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