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Alex777 [14]
3 years ago
5

Consider the following reaction. SO2Cl2 → SO2 + Cl2. After collecting experimental data you found that plotting ln[SO2Cl2] vs. t

ime (s) fit a straight line with the following formula. y = -0.000290t-2.30 Using this information, what is the concentration of SO2Cl2 at 600 s?
Chemistry
1 answer:
Nikitich [7]3 years ago
4 0

Answer:

[SO_2Cl_2]_{600}= 0.0842 M

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 SO_2Cl_2.

The linear equation has the following terms:

y = -0.000290t - 2.30

It is a linear form of the integrated first-order law equation:

ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o

Therefore, the rate constant, k, is:

k = 0.000290 s^{-1}

The natural logarithm of initial molarity is:

ln[SO_2Cl_2]_o = -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

Take the antilog of both sides to find the actual molarity:

[SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M

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