## Slope of k

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

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

### Slope of k

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

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

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

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