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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?

a. 4.50 x 10-4 mol L-l s-l b. 2.25 X 10-4 mol L-l s-l c. 9.00 X 10-4 mol L-l s-l
d. 2.12 x 10-4 mol L-l s-l e. 2.03 x 10-3 mol L-l s-l


  • chemistry - ,

    rate of rxn = 1/2*d(NOBr)/dt = -d(Br2)/dt = -1/2*d(NO)/dt = 4.5E-4

  • chemistry - ,

    I get

    4.50 NOBr x 1Br/2NOBR = 2.25BR

    NOBr cancel out and am left with 2.25BR

    Is this correct?

  • chemistry - ,

    yes that's right

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