Homework 15.23

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Beza Ayalew 1I
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Homework 15.23

Postby Beza Ayalew 1I » Tue Mar 06, 2018 9:54 pm

Can someone explain to me how to do 15.23 part c? Specifically why you would subtract by .153 when you convert the change in concentration of B to moles per liter of A?

Chem_Mod
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Re: Homework 15.23

Postby Chem_Mod » Tue Mar 06, 2018 10:17 pm

You want to find how much of A was left after the reaction. As 0.034 molB was formed and as the ratio of A:B = 2:1, you know that 0.034*2 = 0.068mol A has reacted. This means that 0.153-0.068 = 0.085A was left = [a]t

Maeve Gallagher 1J
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Re: Homework 15.23

Postby Maeve Gallagher 1J » Tue Mar 06, 2018 10:22 pm

The initial concentration of A is .153. The question is asking for the rate constant expressed for the rate constant of A, so you convert .034M B into units of A by multiplying by the mol ratio, 2molA/1molB. This gives you that the change in concentration of A is .068M. To find the final concentration of A, you subtract your initial A, .153 minus your change, .068. This gives you the final concentration of A which you can then plug into the rate equation.

Hubert Tang-1H
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Re: Homework 15.23

Postby Hubert Tang-1H » Tue Mar 06, 2018 10:27 pm

We also know that because the reaction is a first order reaction, we can utilize the equation:
ln [A] = ln [A_0] - kt, and plug in the equations for A_0 = .153 mol/L and A=.153-2*.034=.085 mol/L, and t=115 s


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