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

  • question_answer
    The difference between the heats of reaction at constant pressure and a constant volume for the reaction \[2{{C}_{6}}{{H}_{6\left( 1 \right)}}+15{{O}_{2(g)}}\xrightarrow{{}}12C{{O}_{2(g)}}+6{{H}_{2}}{{O}_{(I)}}\] in Kj is :

    A) - 7.43

    B) + 3.72

    C) - 3.72

    D) + 7.43

    Correct Answer: A

    Solution :

    [a] \[\Delta H=\Delta E+\Delta {{n}_{g}}RT\]
    Here \[\Delta {{n}_{g}}=12-15=-3\]
    Thus,
    \[\Delta H-\Delta E=-3\times 8.314\times 298=-\text{ }7.43\text{ }kJ\]


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