1)The rate law for reaction is = Rate = k[A]²[B]⁰
2)rate constant = 4 × 10⁻³ m/s
3)rate of reaction = 0.01 × 10⁻⁴
1) A + B ------> C + D
Rate = k[A]^x[B]^y
in 1st and 2nd [A] is constant [B] is increased by
0.2 / 0.1 = 2 and initial rate is same.
this means the order with respect to [B] is zero
now, in 1st and 2nd case the [B] is constant and [A] is increased by 0.2 / 0.1 = 2
and the rate has increased by 4. as 16 × 10⁻⁵ / 4 × 10⁻⁵ = 4
this means the order with respect to A must be 2 as 2² = 4
The overall rate expression is
Rate = k[A]²[B]⁰
[B]⁰ = 1
Rate = k[A]²
2) the rate constant k = Rate / [A]²[B]⁰
from the data , k = 4 × 10⁻⁵ / [0.1]²
= 4 × 10⁻³ m/s
3) rate of reaction when [A] = 0.050 M and [B] = 0.100 M
Rate = k [A]²
Rate = 4 × 10⁻³ × [0.050]²
= 1 × 10⁻⁶
Thus, 1)The rate law for reaction is = Rate = k[A]²[B]⁰
2)rate constant = 4 × 10⁻³ m/s
3)rate of reaction = 0.01 × 10⁻⁴
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