The rate of disappearance of hbr in the gas phase reaction 2hbr(g)→h2(g)+br2(g) is 0.140 m s-1 at 150°c. the rate of appearance of h2 is m s-1.

The reaction rate is the rate at which reactants convert to the product. The rate of appearance of hydrogen is

The given reaction,

2 moles of HBr degrade to form 1 mole of Hydrogen and 1 mole of bromine.

The rate of disappearance of HBr = 0.140 m

Since the molar ratio between HBr and Hydrogen is  2:1.

So,

Where,

- Rate of appearance of Hydrogen

- Rate of disappearance of HBr

Put the value,

Therefore, the rate of appearance of hydrogen is .

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answer: your mm

explanation: