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Half life period of 2nd order reaction is
- a)proportional to initial concentration of reactants
- b)independent of initial concentration of reactants
- c)inversely proportional to initial concentration of reactants
- d)inversely proportional to square of initial concentration of reactants
Correct answer is option 'C'. Can you explain this answer?
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The half-life period of a second-order reaction is inversely proportional to the initial concentration of reactants. This means that as the initial concentration of reactants increases, the half-life period decreases.
Explanation:
A second-order reaction is one in which the rate of reaction is proportional to the square of the concentration of the reactants. It can be represented by the following generic equation:
A + B → Products
The rate equation for a second-order reaction is given by:
Rate = k[A][B]
Where [A] and [B] are the concentrations of reactants A and B, and k is the rate constant.
The half-life period of a reaction is the time it takes for the concentration of a reactant to decrease to half of its initial value. In the case of a second-order reaction, the half-life period can be calculated using the integrated rate equation:
t1/2 = 1 / (k[A]0[B]0)
Where t1/2 is the half-life period, [A]0 and [B]0 are the initial concentrations of reactants A and B, and k is the rate constant.
From the equation, we can see that the half-life period is inversely proportional to the product of the initial concentrations of reactants [A]0 and [B]0. This means that as the initial concentration of reactants increases, the denominator of the equation becomes larger, resulting in a smaller value for the half-life period. Conversely, as the initial concentration of reactants decreases, the denominator becomes smaller, resulting in a larger value for the half-life period.
Therefore, the correct answer is option 'C' - the half-life period of a second-order reaction is inversely proportional to the initial concentration of reactants.
Explanation:
A second-order reaction is one in which the rate of reaction is proportional to the square of the concentration of the reactants. It can be represented by the following generic equation:
A + B → Products
The rate equation for a second-order reaction is given by:
Rate = k[A][B]
Where [A] and [B] are the concentrations of reactants A and B, and k is the rate constant.
The half-life period of a reaction is the time it takes for the concentration of a reactant to decrease to half of its initial value. In the case of a second-order reaction, the half-life period can be calculated using the integrated rate equation:
t1/2 = 1 / (k[A]0[B]0)
Where t1/2 is the half-life period, [A]0 and [B]0 are the initial concentrations of reactants A and B, and k is the rate constant.
From the equation, we can see that the half-life period is inversely proportional to the product of the initial concentrations of reactants [A]0 and [B]0. This means that as the initial concentration of reactants increases, the denominator of the equation becomes larger, resulting in a smaller value for the half-life period. Conversely, as the initial concentration of reactants decreases, the denominator becomes smaller, resulting in a larger value for the half-life period.
Therefore, the correct answer is option 'C' - the half-life period of a second-order reaction is inversely proportional to the initial concentration of reactants.
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Half life period of 2nd order reaction isa)proportional to initial concentration of reactantsb)independent of initial concentration of reactantsc)inversely proportional to initial concentration of reactantsd)inversely proportional to square of initial concentration of reactantsCorrect answer is option 'C'. Can you explain this answer?
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