A 1.0 L vessel is charged with 1.00 atm of A, and the following reaction is allowed to come to equilibrium at 298 K.
with \(K_p = 3.10\).
Solution
Part a:
First, we can use an ICE[1] table to solve part a).
|
A |
2 B |
Initial |
1.00 atm |
0 |
Change |
-x |
+2x |
Equilibrium |
1.00 atm - x |
2x |
So (for convenience, consider \(K_p\) to have units of atm)
\[ 3.10 \,atm = \dfrac{(2x)^2}{1.00 \,atm - x} \nonumber \]
Solving for \(x\) yields values of
\[x_1= -1.349 \,atm \nonumber \]
\[x_1= 0.574 \,atm \nonumber \]
Clearly, \(x_1\), while a solution to the mathematical problem, is not physically meaningful since the equilibrium pressure of B cannot be negative. So the equilibrium partial pressures are given by
\[ p_A = 1.00 \,atm - 0.574\, atm = 0.426 \,atm \nonumber \]
\[ p_B = 2(0.574 \,atm) = 1.148 \,atm \nonumber \]
So the mole fractions are given by
\[\ chi_A = \dfrac{0.426 \,atm}{0.426\,atm + 1.148\,atm} = 0.271 \nonumber \]
\[ \chi_B=1-\chi_A = 1-0.271 = 0.729 \nonumber \]
Part b:
The volume is doubled. Again, an ICE table is useful. The initial pressures will be half of the equilibrium pressures found in part a).
|
A |
2 B |
Initial |
0.213 atm |
0.574 atm |
Change |
-x |
+2x |
Equilibrium |
0.213 atm - x |
0.574 atm + 2x |
So the new equilibrium pressures can be found from
\[ 3.10 \,atm = \dfrac{(0.574\,atm + 2x)^2}{0.213\,atm - x} \nonumber \]
And the values of \(x\) that solve the problem are
\[x_1= -1.4077 \,atm \nonumber \]
\[x_1= 0.05875 \,atm \nonumber \]
We reject the negative root (since it would cause both of the partial pressures to become negative. So the new equilibrium partial pressures are
\[ p_A = 0.154\, atm \nonumber \]
\[ p_B = 0.0692\, atm \nonumber \]
And the mole fractions are
\[\chi_A = 0.182 \nonumber \]
\[\chi_B = 0.818 \nonumber \]
We can see that the mole fraction of \(A\) decreased and the mole fraction \(B\) increased. This is the result expected by Le Chatlier’s principle since the lower total pressure favors the side of the reaction with more moles of gas.
Part c:
We introduce 1.000 atm of an inert gas. The new partial pressures are
\[p_A = 0.154 \,atm \nonumber \]
\[p_B = 0.692 \,atm \nonumber \]
\[p_{Ar} = 1.000\, atm \nonumber \]
And because the partial pressures of A and B are unaffected, the equilibrium does not shift! What is affected is the composition, and so the mole fractions will change.
\[ \chi_A = \dfrac{0.154 \,atm}{0.154 atm + 0.692 \,atm + 1.000\, atm} = 0.08341 \nonumber \]
\[ \chi_B = \dfrac{0.692 \,atm}{0.154 atm + 0.692 \,atm + 1.000\, atm} = 0.08341 \nonumber \]
\[ \chi_{Ar} = \dfrac{1.000\, atm}{0.154 atm + 0.692 \,atm + 1.000\, atm} = 0.08341 \nonumber \]
And since
\[ K_p = K_x(p_{tot}) \nonumber \]
\[\dfrac{(0.3749)^2}{0.08342} (1.846\,atm) = 3.1 \nonumber \]
Within round-off error, the value obtained is the equilibrium constant. So the conclusion is that the introduction of an inert gas, even though it increases the total pressure, does not induce a change in the partial pressures of the reactants and products, so it does not cause the equilibrium to shift.