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. $\frac{1}{4}$
• B. 4
• C. $\frac{1}{2}$
• D. 2

Answer 1

Answer: (B)

4

Answer 2

$r_1 = k[A] [B]$ When volume is reduced to half then concentration gets doubled. $\therefore \:\:\: r_2 = k[2AJ] [2B] = 4 k[A] [B] = 4\, r_1$

When the volume of a gaseous reaction vessel is reduced to half, the concentration of the reactants will double.

The rate equation for the reaction is given by r = k[A][B]. If the concentration of both A and B is doubled, the new rate of the reaction can be calculated as:

r' = k[2A][2B] = k(2A)(2B) = 4kAB

So, the new rate of the reaction (r') is four times the initial rate (r). Therefore, the reaction rate relating to the original rate will be 4.

Hence, the correct answer is 4.