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nadya68
10 days ago
11

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 appear

ance 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
Chemistry
1 answer:
lorasvet [2.6K]10 days ago
6 0

Response: The emergence rate of Br_2 is 0.0700Ms^{-1}

Clarification:

According to the rate law, the speed of a reaction is directly linked to the concentration of the reactants, each raised to an experimentally defined stoichiometric exponent known as order.

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

The rate is expressed negatively for reactants because their concentration diminishes over time, whereas it is positively expressed for products as their concentration rises with time.

The rate corresponding to the decrease of HBr = -\frac{1d[HBr]}{2dt}Rate in terms of appearance of [tex]H_2 = \frac{1d[H_2]}{dt}

The rate for the increase of Br_2 = \frac{1d[Br_2]}{dt}

-\frac{1d[HBr]}{2dt}=\frac{d[H_2]}{dt}=\frac{d[Br_2]}{dt}

Provided:

-\frac{1d[HBr]}{dt}=0.140Ms^{-1}

Hence, the emergence rate of Br_2;

\frac{1d[Br_2]}{dt}=-\frac{1d[HBr]}{2dt}=\frac{1}{2}\times 0.140=0.0700Ms^{-1}

Consequently, the emergence rate of Br_2 is 0.0700Ms^{-1}

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