Answer:
[tex][SO_2Cl_2]_{600}= 0.0842 M[/tex]
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 [tex]SO_2Cl_2[/tex].
The linear equation has the following terms:
[tex]y = -0.000290t - 2.30[/tex]
It is a linear form of the integrated first-order law equation:
[tex]ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o[/tex]
Therefore, the rate constant, k, is:
[tex]k = 0.000290 s^{-1}[/tex]
The natural logarithm of initial molarity is:
[tex]ln[SO_2Cl_2]_o = -2.30[/tex]
Using the equation, we may substitute for t = 600 s and obtain the natural logarithm of the concentration at that time:
[tex]ln[SO_2Cl_2]_{600} = -0.000290 s^{-1}\cdot 600 s - 2.30 = -2.474[/tex]
Take the antilog of both sides to find the actual molarity:
[tex][SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M[/tex]
