JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Sample Paper Topic Test - Thermo Chemistry (Thermodynamics)

  • question_answer
    For a gaseous reaction:
    \[2{{A}_{2}}(g)+5{{B}_{2}}(g)\to 2{{A}_{2}}{{B}_{5}}(g)\] at \[{{27}^{o}}C\]
    the heat change at constant pressure is found to be \[-\,50160\,J\]. Calculate the value of internal energy change \[(\Delta E)\]. Given that R = 8.314 J/K mol.

    A) - 34689 J

    B) - 37689 J

    C) - 27689 J

    D) - 38689 J

    Correct Answer: B

    Solution :

    [b] \[2{{A}_{2}}\left( g \right)+5{{B}_{2}}\left( g \right)g\to 2{{A}_{2}}{{B}_{5}}\left( g \right);\Delta H=-50160J\]\[\Delta n=2-\left( 5+2 \right)=-5\,mol\]
    \[\Delta H=\Delta E+{{\left( \Delta n \right)}_{g}}RT\]
    \[-50160=\Delta E+{{\left( \Delta n \right)}_{g}}RT\]
    \[\Delta E=-50160-\left( -5 \right)\left( 8.314 \right)\left( 300 \right)\]

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