Decomposition of h2o2 follows a first order reaction. In fifty minutes...
2H2O2 → 2H2O + O2
In 50 minutes, the concentration is reduced to one fourth
(∵ 0.5/0.125 = 4)
Hence 2 half life periods correxponds to 50 minutes.
t(1/2) = 25 min
The rate constant k = 0.693/t(1/2)
= 0.693/25
Rate of decomposition of H2 x O2 = k[H2O2] = 0.693/25 x 0.05 = 1.39 x 10^−3 mol/min
Rate of formation of oxygen is one half the rate of decomposition of H2O2 .
It is 1/2 x 1.39 x 10^−3 = 6.93 x 10^−4 mol/min
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Decomposition of h2o2 follows a first order reaction. In fifty minutes...
Decomposition of H2O2
Introduction: H2O2 is a chemical compound that decomposes into water and oxygen. This decomposition reaction occurs at different rates depending on various factors such as temperature and concentration of the reactants.
First-Order Reaction: The decomposition of H2O2 follows a first-order reaction, which means that the rate of reaction is directly proportional to the concentration of the reactant.
Given Data: In fifty minutes, the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be calculated.
Calculations:
- To determine the rate constant of the reaction, we can use the formula for first-order reaction: k = (-1/t) x ln([H2O2]t/[H2O2]0), where t is the time, [H2O2]t is the concentration of H2O2 at time t, [H2O2]0 is the initial concentration of H2O2, and ln is the natural logarithm.
- Using the given data, we can calculate the rate constant k: k = (-1/50) x ln(0.125/0.5) = 0.0280 min^-1.
- When the concentration of H2O2 reaches 0.05 M, we can use the first-order rate equation to calculate the rate of formation of O2: rate = k[H2O2] = 0.0280 x 0.05 = 6.93 x 10^-4 M/min.
Conclusion: When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be 6.93 x 10^-4 M/min.
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