Untitled - Kelly Walsh High School

212 CHEMISTRY FOR THE UTTERLY CONFUSED
Don’t Forget!
The problem gives us: P CO 1.22 atm, and P CO2 0.780 atm. The next step is
to enter the given values into the equilibrium constant expression and enter the
values into your calculator.
P2 CO
PCO2 (1.22) 2
Kp 1.9082 1.91
(0.780)
Calculate K c for the preceding equilibrium at 25°C.
This is one of the few problems where an equilibrium expression is not necessary.
It is also one of the few cases where the temperature is important.
We will begin this problem by writing the equation that relates K p and K c:
K p = K c [RT] n g
The “g” subscript, which may not appear in this equation in your textbook, is a
reminder to focus only on the mole of gas present in the equilibrium reaction.
We have the following values given to us: K p 1.91, T 25°C, and R (from
Chapter 5 on gas laws) = (0.0821 Latm/molK). The presence of the kelvin temperature
unit in the value for R should serve as a reminder to convert °C to K.
The temperature is 298 K.
We need to determine the value of n g. We find this value from the reaction:
C(s) CO 2(g) L 2 CO(g). We ignore the solid. There is 1 mol of gas on the
reactant side and 2 mol of gas on the product side, therefore, n g (2 1).
We can now enter these values into the equation to get:
1.91 Kc B¢
1)
0.0821 L·atm
≤ (298 K)R(2
mol·K
K c 0.07806816 0.0781
We will finish this section by examining one of the most common types of equilibria
problems.