## 4.17 C

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

705121606
Posts: 68
Joined: Wed Sep 18, 2019 12:17 am

### 4.17 C

Th problem: 2SO2 (g) + O2 (g) --> 2SO3 (g) at 25C and 1 atm (with piston and constant pressure). initially .030 mol So2 and .030 mol O2 are present in a cylinder. He then adds a catalyst to initiate the reaction.

In part C it tells us the creation goes to completion and the pressure at the reaction remains constant, what is the final volume of the cylinder (include any excess reactant.

I know we can find the final volume using PV=nRT but how do we get the number of moles?
The solutions say the total number of moles remaining in the container will be .030 mol SO3 + 0.15 mol O2 = 0.045 mol of gas at the end of the reaction
i thought that was the moles used up? I don't really understand why the final number of moles would be 0.045

Andrew Liang 1I
Posts: 105
Joined: Fri Aug 30, 2019 12:18 am

### Re: 4.17 C

The total number of moles in the system will be 0.045 because 0.030 mole of product 2SO3 and 0.015 mole of left over O2. Between 0.030 mole of 2SO2 and 0.030 mole of O2, we figure out that 2SO2 is the limiting reagent and O2 is the excessive. After the reaction is complete, all 0.030 mole of 2SO2 will be used up and only 0.015 mole of O2 will be used to form 2SO3. The left over O2 is 0.015 mole. Adding the left over O2 and the amount of 2SO3 produced you get 0.045 total mole in the system.