Question

Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of each of the reactants and products. b. In the first 25.0 s of this reaction, the concentration of HBr drops from 0.600 M to 0.512 M. Calculate the average rate of the reaction during this time interval.c. If the volume of the reaction vessel in part b is 1.50 L, what amount of Br2 (in moles) forms during the first 15.0 s of the reaction?

Answer 1

Answer :

(A) The rate expression will be:

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

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}$

Part A:

The rate expression will be:

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

Part B:

$\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}$

$\text{Average rate}=0.00176M/s$

The average rate of the reaction during this time interval is, 0.00176 M/s

Part C:

As we are given that the volume of the reaction vessel is 1.50 L.

$\frac{d[Br_2]}{dt}=0.00176M/s$

$\frac{d[Br_2]}{15.0s}=0.00176M/s$

$[Br_2]=0.00176M/s\times 15.0s$

$[Br_2]=0.0264M$

Now we have to determine the amount of Br₂ (in moles).

$\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}$

$\text{Moles of }Br_2=0.0264M\times 1.50L$

$\text{Moles of }Br_2=0.0396mol$

The amount of Br₂ (in moles) formed is, 0.0396 mol