Question

The rate of disappearance of HBr in the gas phase reaction 2HBr(g)→H2(g)+Br2(g) is 0.360 Ms−1 at 150 ∘C. The rate of appearance of Br2 is ________ Ms−1. The rate of disappearance of in the gas phase reaction is 0.360 at 150 . The rate of appearance of is ________ . 1.39 0.600 0.720 0.180 0.130

Answer 1

Answer : The rate of appearance of Br₂ is, 0.180 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

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

The expression for rate of reaction will be :

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

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

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

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

The expression for rate of reaction :

The expression for rate of reaction :

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

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

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

As we are given:

$\text{Rate of disappearance of }HBr=0.360M/s$

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

$\frac{d[Br_2]}{dt}=\frac{1}{2}\times 0.360M/s$

$\frac{d[Br_2]}{dt}=0.180M/s$

Thus, the rate of appearance of Br₂ is, 0.180 M/s

Answer 2

The rate of appearance of Br2 in the system is   0.180  Ms−1.

The equation of the reaction is given as;

2HBr(g) → H2(g) + Br2(g)

We know that;

Rate of disappearance of HBr = -1/2d[HBr]dt

Rate of appearance of Br2 = d[Br2]/dt

From the question, we have the rate of disappearance of HBr as 0.360 Ms−1.

Hence;

1/2( 0.360 ) = d[Br2]/dt =  0.180  Ms−1

The rate of appearance of Br2 in the system is   0.180  Ms−1.

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