## Textbook question 7B.15

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Paige Lee 1A
Posts: 136
Joined: Sat Sep 07, 2019 12:16 am

### Textbook question 7B.15

Could someone please explain how to get part c, 10.9 grams? I got that 0.777mols/L remain, but I'm having trouble converting this to grams

Sulfuryl chloride, SO2Cl2, decomposes by first-order kinetics, and kr 5 2.81 3 1023 min21 at a certain temperature. (a) Determine the half-life for the reaction. (b) Determine the time needed for the concentration of SO2Cl2 to decrease to 10% of its initial concentration. (c) If 14.0 g of SO2Cl2 is sealed in a reac- tion vessel of volume 2500. L and heated to the specified tempera- ture, what mass will remain after 1.5 h?

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: Textbook question 7B.15

so we know k=2.8E-3 min-1

1.5 hours=90 minutes

let's convert 14.0 grams to moles
14.0 g / (135g/mol) = .1037 mol, then divide by total liters to get an initial molarity of .000041481....
seeing that your final molarity is so high, I would go back and check this first step with your work too see where you went wrong.

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: Textbook question 7B.15

Now that we have [A]0, we can plug it into out first order reaction equaition

[A]t = [A]0*e^-kt
[A]t = [.00004148]e^-(2.8E-3min-)(90min)

[A]t = .0000322M

now you convert this back to grams...
.000032mol/L * 2500L * (135g/mol) = 10.8grams