How to find the rate constant of a second order reaction?
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Second Order Reactions are characterized by the property that their rate is proportional to the product of two reactant concentrations (or the square of one concentration). Suppose that A ---> products is second order in A, or suppose that A + B ---> products is first order in A and also first order in B. Then the differential rate laws in these two cases are given by Differential Rate Laws:
d[A]/dt = -k [A]2 (for 2A ---> products)
or dx/dt = -k [A][B] (for A + B ---> products)
In mathematical language, these are first order differential equations because they contain the first derivative and no higher derivatives. A chemist calls them second order rate laws because the rate is proportional to the product of two concentrations. By elementary integration of these differential equations Integrated Rate Laws can be obtained:
1/[A] - 1/[A]0 = k t (for 2A ---> products)
or (1/(a-b)) [ln((a-x)/(b-x))-ln(a/b)] = k t (for A + B ---> products)
where a and b are the initial concentrations of A and B (assuming a not equal to b), andx is the extent of reaction at time t. Note that the latter can also be written:
(a-x)/(b-x) = (a/b)exp[(a-b)kt].
A common way for a chemist to discover that a reaction follows second order kinetics is to plot 1/[A] versus the time in the former case, or ln(b(a-x)/a(b-x) versus t in the latter case.
Data Analysis: 1/[A] = 1/[A]0 + k t
A plot of 1/[A] versus t is a straight line with slope k.
d[A]/dt = -k [A]2 (for 2A ---> products)
or dx/dt = -k [A][B] (for A + B ---> products)
In mathematical language, these are first order differential equations because they contain the first derivative and no higher derivatives. A chemist calls them second order rate laws because the rate is proportional to the product of two concentrations. By elementary integration of these differential equations Integrated Rate Laws can be obtained:
1/[A] - 1/[A]0 = k t (for 2A ---> products)
or (1/(a-b)) [ln((a-x)/(b-x))-ln(a/b)] = k t (for A + B ---> products)
where a and b are the initial concentrations of A and B (assuming a not equal to b), andx is the extent of reaction at time t. Note that the latter can also be written:
(a-x)/(b-x) = (a/b)exp[(a-b)kt].
A common way for a chemist to discover that a reaction follows second order kinetics is to plot 1/[A] versus the time in the former case, or ln(b(a-x)/a(b-x) versus t in the latter case.
Data Analysis: 1/[A] = 1/[A]0 + k t
A plot of 1/[A] versus t is a straight line with slope k.
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