Question

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.

Answer 1

Answer: 0.070 m/s

Explanation:

1) balanced chemical equation:

given: 2HBr(g) → H2 (g)+Br2(g)

2) Mole ratios:

2 mol HBr : 1 mol H2

3) That means that every time  2 moles of HBr disappear 1 mol of H2 appears.

That is, the H2 appears at half rate than the HBr disappears.

∴ rate of appearance of H2 = rate of disappearance of HBr / 2 = 0.140 m/s / 2 =  0.070 m/s, which is the answer.