This question is incomplete, the complete question is;
The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.
how long will it take the concentration of A to drop from 0.75 to 0.25M ?
from the unit of the rate constant we know it is a second reaction order
OPTIONS
a) 2.2×10^−3 min
b) 5.5×10^−3 min
c) 180 min
d) 440 min
e) 5.0×10^2 min
Answer:
it will take 440 min for the concentration of A to drop from 0.75 to 0.25M
Option d) 440 min is the correct answer
Step-by-step explanation:
Given that;
Rate constant K = 6.00 × 10⁻³ L.mol⁻¹min⁻¹
3A → 4B
given that it is a second reaction order;
k = 1/t [ 1/A - 1/A₀]
kt = [ 1/A - 1/A₀]
t = [ 1/A - 1/A₀] / k
K is the rate constant(6.00 × 10⁻³)
A₀ is initial concentration( 0.75 )
A is final concentration(0.25)
t is time required = ?
so we substitute our values into the equation
t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)
t = 2.6666 / (6.00 × 10⁻³)
t = 444.34 ≈ 440 min {significant figures}
Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M
Option d) 440 min is the correct answer