# Finding the partial pressure

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.

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...

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