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: 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.

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priyankatutor priyankatutor · Answer 2 · 2021-11-23T08:46:09+01:00

The reaction rate is the rate at which reactants convert to the product. The rate of appearance of hydrogen is [tex]\bold{ 0.070 ms^-^1}[/tex]

The given reaction,

[tex]\bold { 2HBr(g) \rightarrow H_2 (g)+Br_2(g)}[/tex]

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 [tex]\bold {s^-^1}[/tex]

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

So,

[tex]\bold{Ra(H_2) = \dfrac {Rd(HBr)} {2}}[/tex]

Where,

[tex]\bold{Ra(H_2)}[/tex] - Rate of appearance of Hydrogen

[tex]\bold{ Rd(HBr)}[/tex]- Rate of disappearance of HBr

Put the value,

[tex]\bold{Ra(H_2) = \dfrac {Rd(HBr)} {2}}\\\\\bold {Ra(H_2) = \dfrac {0.140 m/s} {2}}\\\\\bold { Ra(H_2) = 0.070 ms^-^1}[/tex]

Therefore, the rate of appearance of hydrogen is [tex]\bold{ 0.070 ms^-^1}[/tex].

To know more about the rate of reaction,

https://brainly.com/question/3334857