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

Hint: 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.