A reaction A→ B follows a second order kinetic. Doubling the concentration of A will increase the rate of formation of B by a factor of:

Question

A reaction A→ B follows a second order kinetic. Doubling the concentration of A will increase the rate of formation of B by a factor of: (A) 1/4 (B) 4 (C) 1/2 (D) 2

Tardigrade Admin · Accepted Answer

For second order reaction r=k[A]2 ... (i) When concentration of A (reactant) is doubled r=k[2A]2 r=4k [A]2 ,,, (ii) Comparing Eqs. (i) and (ii) r=4 r

Q. A reaction $A \to B$ follows a second order kinetic. Doubling the concentration of A will increase the rate of formation of B by a factor of:

1/4

4

1/2

2

For second order reaction
$r = k [A]^{2}$ ... (i)
When concentration of A (reactant) is doubled
$r = k [2 A]^{2}$
$r = 4 k [A]^{2}$ ,,, (ii)
Comparing Eqs. (i) and (ii) $r = 4 r$

Q. A reaction A→B follows a second order kinetic. Doubling the concentration of A will increase the rate of formation of B by a factor of:

1/4

4

1/2

2

For second order reaction r=k[A]2 ... (i) When concentration of A (reactant) is doubled r=k[2A]2 r=4k[A]2,,, (ii) Comparing Eqs. (i) and (ii) r=4r

Q. A reaction $A \to B$ follows a second order kinetic. Doubling the concentration of A will increase the rate of formation of B by a factor of:

For second order reaction
$r = k [A]^{2}$ ... (i)
When concentration of A (reactant) is doubled
$r = k [2 A]^{2}$
$r = 4 k [A]^{2}$ ,,, (ii)
Comparing Eqs. (i) and (ii) $r = 4 r$