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So I saw this problem but can't seem to figure out how to get the answer. I got a test tomorrow lol.

41) $\ce{SO2Cl2}$ decomposes in the gas phase by the following reaction:

$$\ce{SO2Cl2 -> SO2(g) + Cl2(g)}$$

The reaction is first order in $\ce{SO2Cl2}$ and the rate constant is $\ce{3.0 x 10}$^-6 at $600k$.A vessel is charged with 3.3 atm of $\ce{SO2Cl2}$ at 600k. The partial pressure of $\ce{SO2}$ at $\ce{3.0 x 10^5}$ s is ______ atm.

Answer is 2


What I tried:

Starting with PV = nRT(rearranging it to solve for m/v), and using the integrated rate law for a first order equation, I found the concentration of $\ce{Cl2}$ and $\ce{SO4}$. However, I don't know how to isolate the amount of moles to create a mole fraction.

$\ce{ln[SO2Cl2]_i = ln[SO2Cl2]_f -(3.0 * 10^{-6})(3.0 * 10^5)}$

$\ce{[SO2Cl2]_i = \frac{MP}{RT}}$

$\ce{[SO2Cl2]_i = 9.046}$

Plugging back into the first equation:

$\ce{[SO2Cl2]_f = 3.67805}$

$\ce{\frac{m_{Cl}_2 + m_{SO}_2}{V}} = 5.36849$

Not sure where to go from here...

Appreciate your help!

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  • $\begingroup$ Why would you need mole fractions (or moles, for that matter)? $\endgroup$ Jan 16 '20 at 5:48
  • $\begingroup$ @IvanNeretin I guess you don't necessarily need it, but not sure how to find the relative partial pressure of each otherwise. $\endgroup$ Jan 16 '20 at 11:18
  • $\begingroup$ Why, the pressure of SO2Cl2 just decreases exponentially, and that of SO2 grows accordingly. $\endgroup$ Jan 16 '20 at 11:20
  • $\begingroup$ @IvanNeretin How do you know it decreases exponentially? Is there a relation between order and pressure change? or concentration and pressure change? $\endgroup$ Jan 16 '20 at 11:44
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    $\begingroup$ There is a relation between concentration and pressure; that's quite enough. $\endgroup$ Jan 16 '20 at 11:47

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