Consider the following equilibrium:
[tex]2 NOCl(g) \rightleftharpoons 2 NO(g)+Cl_2(g) \quad \Delta G^0=41 . kJ[/tex]
Now suppose a reaction vessel is filled with 8.06 atm of nitrosyl chloride ([tex]NOCl[/tex]) and 7.44 atm of chlorine ([tex]Cl _2[/tex]) at [tex]1171 .{ }^{\circ} C[/tex]. Answer the following questions about this system:
Under these conditions, will the pressure of [tex]Cl _2[/tex] tend to rise or fall?
Is it possible to reverse this tendency by adding NO? In other words, if you said the pressure of [tex]Cl _2[/tex] will tend to rise, can that be changed to a tendency to fall by adding NO? Similarly, if you said the pressure of [tex]Cl _2[/tex] will tend to fall, can that be changed to a tendency to rise by adding NO?
If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO needed to reverse it. Round your answer to 2 significant digits.
Answer (2)
The pressure of C l 2 will tend to rise because Q p < K p . Adding NO can reverse this tendency. The minimum pressure of NO needed to reverse the tendency is approximately 0.54 atm .
;
Calculate the equilibrium constant K p using Δ G 0 = − RT ln K p , resulting in K p = 0.0329 .
Calculate the reaction quotient Q p using initial pressures, which gives Q p = 0 .
Since Q p < K p , the reaction shifts right, and the pressure of C l 2 tends to rise; adding NO can reverse this.
Calculate the minimum pressure of NO needed to reverse the tendency by setting Q p = K p and solving for P NO , resulting in 0.54 atm.
Explanation
Problem Analysis We are given the equilibrium reaction 2 NOCl ( g ) ⇌ 2 NO ( g ) + C l 2 ( g ) with Δ G 0 = 41. k J . The initial pressures of NOCl and C l 2 are 8.06 a t m and 7.44 a t m , respectively, at a temperature of 1171. ∘ C . We need to determine if the pressure of C l 2 will tend to rise or fall, whether adding NO can reverse this tendency, and if so, the minimum pressure of NO needed to reverse it.
Calculating Equilibrium Constant First, we calculate the standard equilibrium constant K p using the formula Δ G 0 = − RT ln K p , where R = 8.314 J / ( m o l ⋅ K ) and T = 1171. ∘ C = 1171 + 273.15 = 1444.15 K . Thus,
ln K p = − RT Δ G 0 = − 8.314 J / ( m o l ⋅ K ) × 1444.15 K 41000 J = − 3.415
K p = e − 3.415 = 0.0329
Calculating Reaction Quotient Next, we calculate the reaction quotient Q p using the initial pressures: Q p = P NOCl 2 P NO 2 P C l 2 . Initially, P NO = 0 , P NOCl = 8.06 a t m , and P C l 2 = 7.44 a t m . Therefore,
Q p = 8.0 6 2 0 2 × 7.44 = 0
Determining the Shift in Equilibrium Since Q p = 0 < K p = 0.0329 , the reaction will shift to the right, favoring the products. This means the pressure of C l 2 will tend to rise.
Reversing the Tendency Since the pressure of C l 2 tends to rise, we want to see if adding NO can reverse this tendency. Yes, adding NO can reverse the tendency.
Calculating Minimum Pressure of NO To find the minimum pressure of NO needed to reverse the tendency, we need to find the pressure of NO such that Q p = K p . Let P NO = x . Then,
Q p = P NOCl 2 x 2 P C l 2 = 8.0 6 2 x 2 × 7.44 = K p = 0.0329
Solving for x :
x 2 = 7.44 0.0329 × 8.0 6 2 = 7.44 0.0329 × 64.9636 = 7.44 2.1373 = 0.2872
x = 0.2872 = 0.536 a t m
Final Answer Rounding the answer to 2 significant digits, we get 0.54 a t m .
Examples
Consider an industrial process where NOCl decomposes into NO and C l 2 . By understanding the equilibrium constant and reaction quotient, engineers can manipulate the reaction conditions (e.g., adding NO ) to optimize the production of NO or C l 2 . This ensures efficient use of resources and minimizes waste. The principles of chemical equilibrium are crucial in designing and controlling chemical reactions in various industrial applications, from pharmaceuticals to materials science.
