In a reaction A arrow Products, when start is made from 8.0 × 10-2 M of A, half-life is found to be 120 minute. For the initial concentration 4.0 × 10-2 M, the half-life of the reaction becomes 240 minute. The order of the reaction is:

Question

In a reaction A arrow Products, when start is made from 8.0 × 10-2 M of A, half-life is found to be 120 minute. For the initial concentration 4.0 × 10-2 M, the half-life of the reaction becomes 240 minute. The order of the reaction is: (A) 0 (B) 1 (C) 2 (D) 0.5

Tardigrade Admin · Accepted Answer

From the half life reaction, (( t 1 / 2)1/( t 1 / 2)2)=(( a 2/ a 1)) n -1 (120/140)=((4 × 10-2/8 × 10-2)) n -1 n =2

Q. In a reaction $A \to$ Products, when start is made from $8.0 \times 1 0^{- 2} M$ of A, half-life is found to be $120$ minute. For the initial concentration $4.0 \times 1 0^{- 2} M$ , the half-life of the reaction becomes $240$ minute. The order of the reaction is:

0

1

2

0.5

From the half life reaction,
$\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{2}} = (\frac{a _{2}}{a _{1}})^{n - 1}$
$\frac{120}{140} = (\frac{4 \times 1 0 ^{- 2}}{8 \times 1 0 ^{- 2}})^{n - 1}$
$n = 2$

Q. In a reaction A→ Products, when start is made from 8.0×10−2M of A, half-life is found to be 120 minute. For the initial concentration 4.0×10−2M, the half-life of the reaction becomes 240 minute. The order of the reaction is:

0

1

2

0.5

From the half life reaction, (t1/2​)2​(t1/2​)1​​=(a1​a2​​)n−1 140120​=(8×10−24×10−2​)n−1 n=2

Q. In a reaction $A \to$ Products, when start is made from $8.0 \times 1 0^{- 2} M$ of A, half-life is found to be $120$ minute. For the initial concentration $4.0 \times 1 0^{- 2} M$ , the half-life of the reaction becomes $240$ minute. The order of the reaction is:

From the half life reaction,
$\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{2}} = (\frac{a _{2}}{a _{1}})^{n - 1}$
$\frac{120}{140} = (\frac{4 \times 1 0 ^{- 2}}{8 \times 1 0 ^{- 2}})^{n - 1}$
$n = 2$