In the reaction, A → Products, a plot of 1/[A] vs. time is linear and the slope is equal to 0.056 M−1 s−1. If the initial concentration of A is 0.80 M, how long will it take one-half of the initial amount of A to react?
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Question:
In the reaction, A → Products, a plot of 1/[A] vs. time is linear and the slope is equal to 0.056 M−1 s−1. If the initial concentration of A is 0.80 M, how long will it take one-half of the initial amount of A to react?
Answer:
Answer:
t = 22.32 s
Explanation:
The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.
The concentration vs time graph of zero order reactions is linear with negative slope.
The concentration vs time graph for a first order reactions is a exponential curve. For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.
The concentration vs time graph for a second order reaction is a hyperbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.
Given that:- 1/[A] vs. time is linear which means it follows second order kinetics.
Thus,
Given that slope = k = 0.056 M⁻¹ s⁻¹.
Integrated rate law for second order kinetic is:
Where, is the final concentration = Half of the initial concentration = 0.80 /2 M = 0.40 M
is the initial concentration = 0.80 M
So,
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