Question

The catalytic decomposition of ${{\text{H}}_2}{{\text{O}}_2}$ was studied by titrating it at different intervals with ${\text{KMn}}{{\text{O}}_4}$ and the following data was obtained:

T in seconds	0	600	1200
Volume of ${\text{KMn}}{{\text{O}}_4}$ in mL	$22.8$	$13.8$	$8.3$

Calculate the velocity constant for the reaction assuming it to be a first order reaction.

Answer 1

Using the formula we will first calculate the initial concentration value. Then using the given data we will calculate the rate constant or the rate constant.

Formula used: ${\text{A}} = {{\text{A}}_{\text{o}}}{{\text{e}}^{ - {\text{Kt}}}}$
Here A is the final volume, ${{\text{A}}_{\text{o}}}$ is initial volume and K is the rate constant and t is time.

Complete step by step answer:
We will use the formula for the first order equation. At time equals to zero the initial concentration is $22.8$ , this is what was given to us in the question. Time zero indicates that the reaction has not started yet. Hence the value of ${{\text{A}}_{\text{o}}} = 22.8$
Now we have been given the value of volume at time equals 600. Using time as 600 and the concentration A as $13.8$ and the initial concentration ${{\text{A}}_{\text{o}}} = 22.8$ in the formula we will get:
$13.8 = 22.8 \times {{\text{e}}^{ - 600{\text{t}}}}$
Now we will solve the equation as:
$0.6052 = {{\text{e}}^{ - 600{\text{t}}}}$
Natural log and exponential are reciprocal of each other and hence we will get:
$\ln 0.6052 = \ln {{\text{e}}^{ - 600{\text{t}}}}$
Substituting the log values we will get:
$- 0.50219 = - 600{\text{K}}$
Rearranging this we will get the value of rate constant as:
${\text{K}} = 8.369 \times {10^{ - 4}}{{\text{s}}^ -1 }$

Note: A first order reaction is that reaction in which rate of reaction is directly proportional to the concentration of reactant. For a zero order reaction the rate of reaction is independent of the initial concentration. Similarly for a second order reaction the rate of reaction is directly proportional to the second power of the reactant. Natural log has the base e whereas the log has the base 10. A rate constant is the constant of proportionality and is a function of temperature and nature of reaction.