If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at constant press

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If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at constant pressure at thesame temperature

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  1. Answer:

    The heat at constant pressure is -3,275.7413 kJ

    Explanation:

    The combustion equation is 2C₆H₆ (l) + 15O₂ (g)  → 12CO₂ (g) + 6H₂O (l)

    [tex]\Delta n_g[/tex] = (12 – 15)/2 = -3/2

    We have;

    [tex]\Delta H = \Delta U + \Delta n_g\cdot R\cdot T[/tex]

    Where R and T are constant, and ΔU is given we can write the relationship as follows;

    [tex]H = U + \Delta n_g\cdot R\cdot T[/tex]

    Where;

    H = The heat at constant pressure

    U = The heat at constant volume = -3,272 kJ

    [tex]\Delta n_g[/tex] = The change in the number of gas molecules per mole

    R = The universal gas constant = 8.314 J/(mol·K)

    T = The temperature = 300 K

    Therefore, we get;

    H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

    The heat at constant pressure, H = -3,275.7413 kJ.

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