Question

The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appearance of Br2 is ________ M s-1. The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appearance of Br2 is ________ M s-1.

a. 0.0700
b. 1.28
c. 0.0196
d. 0.280
e. 0.374

Answer 1

Answer: The rate of appearance of
`Br_2` is
`0.0700Ms^(-1)`

Step-by-step explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

`2HBr(g)\rightarrow H_2(g)+Br(g)`

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of HBr =
`-(1d[HBr])/(2dt)</p><p>Rate in terms of appearance of [tex]H_2` =
`(1d[H_2])/(dt)`

Rate in terms of appearance of
`Br_2` =
`(1d[Br_2])/(dt)`

`-(1d[HBr])/(2dt)=(d[H_2])/(dt)=(d[Br_2])/(dt)`

Given :

`-(1d[HBr])/(dt)=0.140Ms^(-1)`

The rate of appearance of
`Br_2`;

`(1d[Br_2])/(dt)=-(1d[HBr])/(2dt)=(1)/(2)* 0.140=0.0700Ms^(-1)`

Thus rate of appearance of
`Br_2` is
`0.0700Ms^(-1)`