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Worked Example : Calculating Equilibrium Partial Pressure
Question: A container at 800°C was filled NOCl gas which decomposes to form NO gas and chlorine gas.
The equilibrium partial pressure of NOCl was 0.657 atm, what was the partial pressure of NO gas?
K_{P} = 1.8 × 10^{2} for the reaction NOCl_{(g)} ⇋ NO_{(g)} + ½Cl_{2(g)}
Solution:
(Based on the StoPGoPS approach to problem solving in chemistry.)
 What is the question asking you to do?
Calculate the partial pressure of NO gas.
P_{(NO(g))} = ? atm
 What data (information) have you been given?
(a) NOCl_{(g)} ⇋ NO_{(g)} + ½Cl_{2(g)}
(b) K_{P} = 1.8 × 10^{2}
(c) P_{(NOCl(g))} = 0.657 atm
 What is the relationship between what you know and what you need to find out?
Write the expression for K_{P}
K_{P} = 
P_{(NO)}P^{½}_{(Cl2)} 

P_{(NOCl)} 
Write expressions for the equilibrium partial pressure of each species based on the reaction stoichiometry (mole ratio):
P_{(NOCl)} = 0.657 atm
P_{(NO)} = x atm
P_{(Cl2)} = ½x atm
 Subsitute the values into the expression for K_{P} and solve for P_{(NO)}:
1.8 × 10^{2} = 
x(½x)^{½} 

0.657 
Simplify the expression for K_{P} and solve for x
1.8 × 10^{2} × 0.657= (½x^{3})^{½}
0.0118 = (½x^{3})^{½}
(0.0118)^{2} = ½x^{3}
1.4 × 10^{4} = ½x^{3}
1.4 × 10^{4} × 2 = x^{3}
2.8 × 10^{4} = x^{3}
^{3}√2.8 × 10^{4} = x
0.065 = x = P_{NO}
 Is your answer plausible?
Check your answer by substituting the values back into the equation:
P_{NOCl} = 0.657 atm
P_{NO} = 0.065 atm
P_{Cl2} = ½ × 0.065 = 0.0325 atm
K_{P} = 
P_{(NO)}P^{½}_{(Cl2)} 

P_{(NOCl)} 
K_{P} = 
0.065(0.0325)^{½} 

0.657 
K_{P} = 1.8 × 10^{2} which is the same as the value we were given in the question so we are confident our answer is correct.
 State your solution to the problem:
P_{(NO)} = 0.065 atm
Worked Example : Converting K_{c} to K_{P}
Question: For the equilibrium:
2NOCl_{(g)} ⇋ 2NO_{(g)} + Cl_{2(g)}
K_{c} = 3.75 × 10^{6} at 796°C.
Calculate K_{P} for this reaction at this temperature.
Solution:
(Based on the StoPGoPS approach to problem solving in chemistry.)
 What is the question asking you to do?
Calculate K_{P}
K_{P} = ?
 What data (information) have you been given?
(a) 2NOCl_{(g)} ⇋ 2NO_{(g)} + Cl_{2(g)}
(b) K_{c} = 3.75 × 10^{6}
(c) T = 796°C
Convert temperature in celsius to kelvin:
T = 796 + 273 = 1069 K
 What is the relationship between what you know and what you need to find out?
Write the expression to convert K_{c} to K_{p}:
K_{P} = K_{c}(RT)^{Δn}
Use data sheet to find value for ideal gas constant, R,
R = 0.0821 (Ideal Gas Constant)
Use balanced chemical equation to calculate Δn :
Δn = (2 + 1)  2 = 1
 Substitute the values into the equation and solve for K_{P}:
K_{P} = K_{c}(RT)^{Δn}
K_{P} = 3.75 × 10^{6}(0.0821 × 1069)^{1}
K_{P} = 3.29 × 10^{4}
 Is your answer plausible?
Use your calculated value of K_{P} and the given value of K_{c} and temperature to find R.
K_{P} = K_{c}(RT)^{Δn}
K_{P}/K_{c} = (RT)^{Δn}
3.29 × 10^{4}/3.75 × 10^{6} = (RT)^{1}
87.73 = RT
87.73 = R(796 + 273)
87.73 = R × 1069
R = 87.73/1069 = 0.0821
Since this value of R is the same as that on the Data Sheet we are reasonably confident our answer is correct.
 State your solution to the problem:
K_{P} = 3.29 × 10^{4}