Slope of k

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Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

Slope of k

Postby Tam To 1B » Mon Mar 11, 2019 10:26 pm

Why is the slope, k, of second order reactions positive?

Nicole Elhosni 2I
Posts: 62
Joined: Fri Sep 28, 2018 12:28 am

Re: Slope of k

Postby Nicole Elhosni 2I » Mon Mar 11, 2019 10:54 pm

When you take the integral of the rate equation, -d[A]/dt=k[A]², as it is a second order reaction and you are assuming all the coefficient a =1. You can use separation of variables and rewrite the equation as (-1/[A]²)(d[A]) = kdt. Integrate both sides and you will get 1/[A]=kt+C.
The antiderivative of -[A]⁻² is [A]⁻¹, which is why the slope of k on the right hand side of the equation will be positive.

Nicole Lee 4E
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am
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Re: Slope of k

Postby Nicole Lee 4E » Mon Mar 11, 2019 10:57 pm

When you integrate 1/[A], you get a negative sign so the slope of k switches from negative to positive.

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