CET Karnataka Medical CET - Karnataka Medical Solved Paper-2004

  • question_answer
    The enthalpy of combustion of methane at \[{{25}^{o}}C\]is 890 kJ. The heat liberated when \[3.2g\]of methane is burnt is air is:

    A)  \[445kJ\]        

    B)  \[278kJ\]

    C)  \[-890kJ\]

    D)  \[178kJ\]

    Correct Answer: D

    Solution :

    The combustion of methane can be represented by the following equation: \[\underset{12+4=16}{\mathop{C{{H}_{4}}}}\,+2{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+2{{H}_{2}}O+890kJ\] \[\because \] \[16gC{{H}_{4}}\] burns in air to liberate \[=890\text{ }kJ\] of heat \[\therefore \] \[3.2g\text{ }C{{H}_{4}}\]will liberate \[=\frac{890\times 3.2}{16}\] \[=178\text{ }kJ\]of heat


You need to login to perform this action.
You will be redirected in 3 sec spinner