Answer:
![[SO_2Cl_2]_{600}= 0.0842 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D_%7B600%7D%3D%200.0842%20M)
Explanation:
Some theoretical knowledge is required here. We should understand that whenever we plot the natural logarithm, ln, of a concentration vs. time and obtain a straight line, this indicates a first-order reaction. That said, since this is the case here, we have a first-order reaction with respect to
.
The linear equation has the following terms:

It is a linear form of the integrated first-order law equation:
![ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o](https://tex.z-dn.net/?f=ln%5BSO_2Cl_2%5D_t%20%3D%20-kt%20%2B%20ln%5BSO_2Cl_2%5D_o)
Therefore, the rate constant, k, is:

The natural logarithm of initial molarity is:
![ln[SO_2Cl_2]_o = -2.30](https://tex.z-dn.net/?f=ln%5BSO_2Cl_2%5D_o%20%3D%20-2.30)
Using the equation, we may substitute for t = 600 s and obtain the natural logarithm of the concentration at that time:
![ln[SO_2Cl_2]_{600} = -0.000290 s^{-1}\cdot 600 s - 2.30 = -2.474](https://tex.z-dn.net/?f=ln%5BSO_2Cl_2%5D_%7B600%7D%20%3D%20-0.000290%20s%5E%7B-1%7D%5Ccdot%20600%20s%20-%202.30%20%3D%20-2.474)
Take the antilog of both sides to find the actual molarity:
![[SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D_%7B600%7D%3De%5E%7B-2.474%7D%20%3D%200.0842%20M)