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J & K CET Medical J & K - CET Medical 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\[-284\text{ }kJ\text{ }mo{{l}^{-1}}\]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 :

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


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