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1. Nitric oxide reacts with bromine gas at elevated temperatures according to the equation,
2 NO(g) + Br2(g) = 2 NOBr(g)

The experimental rate law is rate = k[NO][Br2]. In a certain reaction mixture the rate of formation of NOBr(g) was found to be 4.50 x 10-4 mol L-l s-l. What is the rate of consumption of Br2(g), also in mol L-l s-l?

I think that the answer is 9.00 X e-4 because I multiplied 4.50 by 2.

Is this correct?

  • Chemistry - ,

    Yes it is.

  • Chemistry - ,

    thank you!

  • Chemistry - ,

    Wouldn't you divide it by 2?

    4.50 NOBr x 1Br/2NOBR = 2.25BR

    The NOBr cancel. Is this not correct

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