If the rate of decrease for the partial pressure of N2H4 in a closed reaction vessel is 76 torr/h , what is the rate of change for the parti

Question

If the rate of decrease for the partial pressure of N2H4 in a closed reaction vessel is 76 torr/h , what is the rate of change for the partial pressure of NH3 in the same vessel

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Julia 1 month 2021-10-15T14:04:21+00:00 1 Answer 0 views 0

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  1. Emma
    0
    2021-10-15T14:05:37+00:00

    The reaction is missing and it’s ;

    N2H4 + H2 —> 2NH3

    It asks for the total pressure too.

    Answer:

    A) Rate of change for NH3 = 152 torr/h

    B) The total pressure in the vessel will remain the same.

    Step-by-step explanation:

    N2H4 + H2 —> 2NH3

    1 mole of N2H4 yields 2 moles of NH3.

    From the question, the rate given for N2H4 is 76 torr/h.

    Thus, The rate of change for NH3 will be = 2 x 76 = 152 torr/h

    Now, on the reaction side, 1 mole of N2H4 reacts with 1 mole of H2. So we have 2 moles on the left hand side.

    While on the product side, 2 moles of NH3 are produced.

    So the total pressure will remain the same because for every 2 moles on the reaction side, 2 moles are gotten on the product side.

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