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17 November, 18:51

After 1.60 mol of NH3 gas is placed in a 1600-cm3 box at 25°C, the box is heated to 500 K. At this temperature, the ammonia is partially decomposed to N2 and H2, and a pressure measurement gives 4.85 MPa. Find the number of moles of each component present at 500 K.

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  1. 17 November, 21:19
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    1.3333 mol for ammonia, 0.1334 mol for N₂ and 0.400 mol for H₂

    Explanation:

    To calculate the number of mole in the resulting mixture

    PV = nRT where is gas constant = 8.314 J / mol. K T = 500 K Volume = 1600 cm³ = 1600 / 1000000 = 0.0016 m³

    substitute the values into the equation

    4.85 * 10⁶ Pa * 0.0016 m³ = 8.314 * 500 * n

    7760 = 4157 n

    n = 7760 / 4157 = 1.8667308 mol

    equation of the reaction

    2 NH₃ (g) → N₂ (g) + 3 H₂ (g)

    2 moles of ammonia yields 1 mole of nitrogen, and 3 moles of hydrogen

    1 moles of ammonia will yield 0.5 mole of nitrogen and 1.5 moles of hydrogen

    since the ammonia did not fully decomposed

    y moles of ammonia will yield 0.5y mole of nitrogen and 1.5y moles of hydrogen

    the mole of ammonia remaining = 1.60 - y

    sum of the mole in the box after reaction = (1.60 - y) + 0.5 y + 1.5 y = 1.6 + y = 1.8667308 mol

    y = 1.8667308 - 1.6 = 0.26673 mol

    number of ammonia remaining = 1.6 - 0.26673 mol = 1.3333 mol

    number of N₂ present = 0.5 * 0.26673 mol = 0.1334 mol

    number of H₂ present = 1.5 * 0.26673 mol = 0.400 mol
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