# 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...