Question

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 500 ml flask with 3.4 atm of sulfur dioxide gas and 1.3 atm of oxygen gas at 31. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.52 atm. Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answert significant digits. x I ? Explanation

Answer 1

Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

                                2SO2      +        O2         ⇄              2SO3

Initial                      3.4 atm           1.3 atm                         0 atm

Change                    -2x                    - x                                + 2x

Equilibrium            3.4 - 2x            1.3 -x                          0.52 atm

In order to know the x value:

2x = 0.52

x=(0.52)/2= 0.26

                               2SO2             +          O2              ⇄              2SO3

Equilibrium        3.4 - 0.52                1.3 - 0.26                     0.52 atm

Equilibrium        2.88 atm                 1.04 atm                      0.52 atm

with the partial pressure in the equilibrium, we can obtain Kp.

$Kp=\frac{PSO3^2}{PSO2^2 PO2}=\frac{(0.52)^2}{(2.88)^2(1.04)}=0.03135$