A student ran the following reaction in the laboratory at 1080 K: 2SO3(g) 2SO2(g) + O2(g) When she introduced SO3(g) at a pressure of 0.948 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of SO3(g) to be 0.369 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Kp =

0.078

Explanation:

The equation is :

Initial         0.948                  -----                   ----

Change      -2x                     +2x                  +x

Final          0.369                  2x                    x

So the total pressure must reman same = 0.948

And the total pressure = partial pressure of all gases

0.948   =   ( 0.369 + 2x + x )

0.948  =  0.369 + 3x

  = 0.193 atm

So the partial pressure of  = 0.193 x 2

                                                   = 0.386 atm

Partial pressure of  =  0.193 atm

Therefore,

    = 0.078

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.