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14 July, 16:03

What would happen to the rate of a reaction with rate law rate = k no2 h2 if the concentration of h2 were halved?

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  1. 14 July, 18:25
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    The question is not formatted correctly. The proper question should be

    What would happen to the rate of a reaction with rate law

    rate=k [NO]2[H2]

    if the concentration of H2 were halved?

    Answer:

    Rate would be halved.

    Explanation:-

    Let originally [NO] = 1 and [H2]=1.

    Then original rate = k [1]2 [1] = k.

    Now if [H2] is halved then new [H2] = 1/2

    While [NO] remains same.

    Then

    new rate = k [1]2 [1/2] = k/2

    Thus the rate is halved.
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