A 50-mL Mohr buret is connected to a 1-ft length of capillary tubing. The buret is then filled with water and the volume of water versus time is monitored as the water gradually flows through the capillary tubing.
This process is found to be first-order with a rate constant of 0.014 mL/sec. What will happen if the length of the capillary tube is doubled?
a)The process will still be first-order, but the rate constant will decrease.
b)The order and rate constant for the process will both change.
c)The order of the process will change, but the rate constant will be constant.
d)The process will still be first-order, but the rate constant will increase.
I know the answer's not C for sure
and if there's a choice E then it's not that either
It will still be first order but the constant will decrease.
To determine what happens if the length of the capillary tube is doubled, we need to understand the relationship between the length of the capillary tube and the rate constant in a first-order process.
In a first-order process, the rate of change of a species is directly proportional to its concentration or amount remaining. Mathematically, this can be represented as:
rate = -k[A]
Where rate is the rate of change, k is the rate constant, and [A] is the concentration or amount remaining of the species.
In this scenario, the concentration or amount remaining is represented by the volume of water in the buret. The rate is given as 0.014 mL/sec.
If the length of the capillary tube is doubled, it means the resistance to flow through the capillary tube is increased. This increased resistance would result in a slower rate of water flow.
Since the rate of change is directly proportional to the rate constant, a slower rate of water flow would mean a smaller rate constant. Therefore, the correct answer is (a) The process will still be first-order, but the rate constant will decrease.