## 7B.3

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

RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

### 7B.3

How do you do part c of this question.( i.e. how do you find concentration of A)

Determine the rate constant for each of the following first- order reactions, in each case expressed for the rate of loss of A:

c) 2A > B + C
[A]0 = 0.153 mol/L
after 115s the concentration of B rises to 0.034 mol/L

Posts: 103
Joined: Sat Aug 24, 2019 12:15 am

### Re: 7B.3

You write out the rate law for the first-order reaction, then you find that [A] decreased by .068M over 115 seconds as the rate, then you can plug in all the information you know, leaving the k constant as the only unknown since you know [A], the rate, and that the reaction is first-order, and solve for that.

RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

### Re: 7B.3

Adam Kramer 1A wrote:You write out the rate law for the first-order reaction, then you find that [A] decreased by .068M over 115 seconds as the rate, then you can plug in all the information you know, leaving the k constant as the only unknown since you know [A], the rate, and that the reaction is first-order, and solve for that.

After you subtract .068M from 0.153M you get 0.85M for A. But when I plugged it in, I am still getting the wrong answer. I keep getting 0.0454 1/s when the answer is 5.1x10-3 1/s. Is the solution manual just incorrect?

HuyHa_2H
Posts: 100
Joined: Wed Sep 11, 2019 12:15 am

### Re: 7B.3

You first find the concentration of A after B undergoes the change which is 0.153 M of A - (2 mol of A divided by 1 mol of B)*(0.034 mol of B per liter) which gives you 0.085 M of A afterwards. Then plug this concentration as well as the initial concentration into the rate law of the first order and isolate k where k = ln(0.153M/0.085 M)/115 seconds which should get you 0.0051 seconds in the end.

Sydney Myers 4I
Posts: 100
Joined: Fri Aug 09, 2019 12:17 am

### Re: 7B.3

You have to take into account that the rate of formation of B is not equal to the rate of consumption of A. They are separated by a factor of 2, the coefficient.

Petrina Kan 2I
Posts: 102
Joined: Fri Aug 30, 2019 12:17 am

### Re: 7B.3

Why do we use the first order formula for this problem? Since there is 2A does that mean it is second order, or is that only when there are 2 different reactants?

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