Decomposition of H2O2 follows a first order reaction. 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:
Question
Decomposition of H2O2 follows a first order reaction. 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:
Solution
To solve this problem, we first need to find the rate constant (k) for the reaction. We know that for a first order reaction, the rate constant can be calculated using the formula:
k = 1/t * ln([A]0/[A]t)
where:
- t is the time elapsed (50 minutes in this case)
- [A]0 is the initial concentration (0.5 M in this case)
- [A]t is the concentration after time t (0.125 M in this case)
Substituting these values into the formula, we get:
k = 1/50 * ln(0.5/0.125) k = 0.01832 min^-1
Now, we can use this rate constant to find the rate of formation of O2 when the concentration of H2O2 is 0.05 M. For a first order reaction, the rate of reaction is given by:
Rate = k[A]
where [A] is the concentration of the reactant. Substituting the values we have:
Rate = 0.01832 * 0.05 Rate = 0.000916 M/min
So, the rate of formation of O2 when the concentration of H2O2 is 0.05 M is 0.000916 M/min.
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