The rate of disappearance of hbr in the gas phase reaction 2hbr(
g. → h2(
g. + br2(
g. is 0.301 m s-1 at 150°
c. the rate of appearance of br2 is m s-1.

The correct answer would be 0.1505 M/s. Given the rate of disapperance of HBr, we can easily calculate the rate of apearance of the products by looking at the coefficients of the substances in the reaction. In this case, for every two moles of HBr, 1 mole of Br2 is being produced so the reaction rate would be:

- 1/2 (r(HBr)) = rBr2
- 1/2 (-0.301) = rBr2
Rate of appearance of Br2 = 0.1505 M/s

I’m not sure what your trying to say here.. can you clarify again?

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