The rate equation of a gaseous reaction is given by, r = k[A][B]. If the volume of reaction vessel is suddenly reduced to half of the initial volume, the reaction rate relating to the original rate will be

Question

The rate equation of a gaseous reaction is given by, r = k[A][B]. If the volume of reaction vessel is suddenly reduced to half of the initial volume, the reaction rate relating to the original rate will be (A)  (1/4) (B) 4 (C)  (1/2) (D) 2

Tardigrade Admin · Accepted Answer

r1 = k[A] [B] When volume is reduced to half then concentration gets doubled. ∴ : :: r2 = k[2AJ] [2B] = 4 k[A] [B] = 4 r1

Q. The rate equation of a gaseous reaction is given by, $r = k [A] [B]$ . If the volume of reaction vessel is suddenly reduced to half of the initial volume, the reaction rate relating to the original rate will be

$\frac{1}{4}$

4

$\frac{1}{2}$

2

$r_{1} = k [A] [B]$
When volume is reduced to half then concentration gets doubled.
$∴ r_{2} = k [2 A J] [2 B] = 4 k [A] [B] = 4 r_{1}$

Q. The rate equation of a gaseous reaction is given by, r=k[A][B]. If the volume of reaction vessel is suddenly reduced to half of the initial volume, the reaction rate relating to the original rate will be

41​

4

21​

2

r1​=k[A][B] When volume is reduced to half then concentration gets doubled. ∴r2​=k[2AJ][2B]=4k[A][B]=4r1​

Q. The rate equation of a gaseous reaction is given by, $r = k [A] [B]$ . If the volume of reaction vessel is suddenly reduced to half of the initial volume, the reaction rate relating to the original rate will be

$\frac{1}{4}$

$\frac{1}{2}$

$r_{1} = k [A] [B]$
When volume is reduced to half then concentration gets doubled.
$∴ r_{2} = k [2 A J] [2 B] = 4 k [A] [B] = 4 r_{1}$