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 and rate constant for the process will both be constant.
d)The order of the process will change, but the rate constant will be constant.
e)The process will still be first-order, but the rate constant will increase.
why is it not c or d,
A). the process will still be first order, but the rate constant will decrease.
Joe is right.
i looked at all the other stupid websites and tried D and C. They were both wrong.
A is the correct answer I just tried it right now.
To determine the effect of doubling the length of the capillary tube, we need to understand the relationship between reaction order and rate constant in a first-order process.
In a first-order reaction, the rate of reaction depends on the concentration of a single reactant raised to the power of one. This means that changing the length of the capillary tube, which does not affect the concentration of the reactant (water), will not alter the reaction order. Therefore, option d) can be ruled out.
The rate constant, on the other hand, is a measure of how fast the reaction occurs. It is specific to a particular reaction and is influenced by factors such as temperature, pressure, and the characteristics of the reactants. Any change in these factors can alter the rate constant.
In the given scenario, the rate constant is given as 0.014 mL/sec. Doubling the length of the capillary tube will increase the resistance to flow. As a result, the water will take longer to travel through the longer capillary, causing the rate of flow to decrease. Therefore, option a) is the correct answer: the process will still be first-order, but the rate constant will decrease.