When concentration of reactants in reaction A→ B is increased by 8 times the rate increases only 2 times, the order of the reaction would be

Question

When concentration of reactants in reaction A→ B is increased by 8 times the rate increases only 2 times, the order of the reaction would be (A) 2 (B) 1/3 (C) 4 (D) 1/2

Tardigrade Admin · Accepted Answer

r 1= k [ A ] n r 2= r 1= k [ 8 A ] n ∴ ( 2 r 1/ r 1)=( 8 ) n (2)1=(2)3 n ∴ 3 n =1 n =( 1 / 3 )

Q. When concentration of reactants in reaction A $\to$ B is increased by 8 times the rate increases only 2 times, the order of the reaction would be

2

1/3

4

1/2

$r_{1} = k [A]^{n}$

$r_{2} = r_{1} = k [8 A]^{n}$

$∴ \frac{2 r _{1}}{r _{1}} = (8)^{n}$

$(2)^{1} = (2)^{3 n}$

$∴ 3 n = 1$

$n = \frac{1}{3}$

Q. When concentration of reactants in reaction A→ B is increased by 8 times the rate increases only 2 times, the order of the reaction would be

2

1/3

4

1/2

r1​=k[A]n r2​=r1​=k[8A]n ∴r1​2r1​​=(8)n (2)1=(2)3n ∴3n=1 n=31​

Q. When concentration of reactants in reaction A $\to$ B is increased by 8 times the rate increases only 2 times, the order of the reaction would be

$r_{1} = k [A]^{n}$

$r_{2} = r_{1} = k [8 A]^{n}$

$∴ \frac{2 r _{1}}{r _{1}} = (8)^{n}$

$(2)^{1} = (2)^{3 n}$

$∴ 3 n = 1$

$n = \frac{1}{3}$