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The decomposition of sulfuryl chloride (SO₂Cl₂) is a first-order process. The rate constant for the decomposition at 660 K is 4.5 x 10⁻² s⁻¹.

a) If we begin with an initial composition of 450 torr, what is the pressure of this substance after 1 minute?
b) At what time will the pressure of SO₂Cl₂ decline to 1/10 of its original value?

User Xatyrian
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Final answer:

To calculate the pressure of SO2Cl2 after 1 minute, use the formula [A]t = [A]0 * e^{-kt} with the given parameters. For the time it takes the pressure to decline to 1/10 of its initial value, use the formula t = (1/k) * ln(10) with the provided rate constant.

Step-by-step explanation:

The decomposition of sulfuryl chloride (SO2Cl2) is a first-order process. For a first-order reaction, the integrated rate law can be written as ln([A]t/[A]0) = -kt, where [A]0 is the initial concentration or pressure, [A]t is the concentration or pressure at time t, k is the rate constant, and t is the time.

a) To find the pressure of SO2Cl2 after 1 minute, we can rearrange the integrated rate law to solve for [A]t: [A]t = [A]0 * e-kt. Plugging in our values ([A]0 = 450 torr, k = 4.5 x 10-2 s-1, and t = 60 s), we can calculate the pressure after 1 minute.

b) To find the time when the pressure of SO2Cl2 declines to 1/10 of its original value, we can use the same integrated rate law. We set [A]t = [A]0/10 and solve for t, resulting in t = (1/k) * ln(10).

User Garromark
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