When CO2(g) is put in a sealed container at 730 K and a pressure of 10.0 atm and is heated to 1420 K , the pressure rises to 24.1 atm . Some

Question

When CO2(g) is put in a sealed container at 730 K and a pressure of 10.0 atm and is heated to 1420 K , the pressure rises to 24.1 atm . Some of the CO2 decomposes to CO and O2.

Calculate the mole percent of CO2 that decomposes.
Express your answer using two significant figures.

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Lệ Thu 5 years 2021-07-30T14:05:29+00:00 1 Answers 40 views 0

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  1. Nho
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    2021-07-30T14:06:32+00:00
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    Answer:

    48%

    Explanation:

    Based on Gay-Lussac’s law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:

    P1T2 = P2T1

    Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:

    P1 = 10.0atm

    T2 = 1420K

    P2 = ?

    T1 = 730K

    P2 = 10.0atm*1420K / 730K

    P2 = 19.45 atm

    The CO2 reacts as follows:

    2CO2 → 2CO+ O2

    Where 2 moles of gas react producing 3 moles of gas

    Assuming the 100% of CO2 react, the pressure will be:

    19.45atm * (3mol / 2mol) = 29.175atm

    As the pressure rises just to 24.1atm the moles that react are:

    24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present

    The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:

    0.48mol / 1mol * 100 = 48%

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