Question

Question: At 25 °C and at 1 atm, the partial pressures in an equilibrium mixture of N₂O₄ and NO₂ are PN₂O₄= 0.70 atm and PNO₂ = 0.30 atm.

(a) Predict how the pressures of NO₂ and N₂O₄ will change if the total pressure increases to 9.0 atm. Will they increase, decrease, or remain the same?

(b) Calculate the partial pressures of NO₂ and N₂O₄ when they are at equilibrium at 9.0 atm and 25 °C

Short answer

1. Both gases must increase in pressure.
2. The partial pressures are \({P_{{{\rm{N}}_2}{{\rm{O}}_4}}} = 8\,atm\)and \({P_{{\rm{N}}{{\rm{O}}_2}}} = 1\,atm\).

Step-by-step solution

Define Interpretation:

The partial pressures of NO₂ and N₂O₄ at equilibrium at 9.0 atm and 25(C should be calculated).

The affect (decrease, increase or remain same) in the pressures of NO₂ and N₂O₄ should be determined when the total pressure increases to 9.0 atm.

Determine change in pressure:

The equilibrium mixture has equal rates of the forward and backward (reverse) reactions.

For a chemical reaction as follows:

\(A \to B + C\)

The expression for equilibrium constant will be:

\({K_C} = \frac{{\left( B \right)\left( C \right)}}{{\left( A \right)}}\)

Here, [A], [B] and [C] is equilibrium concentration of A, B and C respectively.

a)

At 25⁰C and at 1 atm, the partial pressures in an equilibrium mixture of N₂O₄ and NO₂ are \({P_{{{\rm{N}}_2}{{\rm{O}}_4}}}\) =0.70 atm And \({P_{{\rm{N}}{{\rm{O}}_2}}}\)=0.30 atm.

The reaction is\(2{\rm{N}}{{\rm{O}}_{2(g)}} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {{\rm{N}}_2}{{\rm{O}}_{4(g)}}\)

At equilibrium,\({K_p} = \frac{{\left( {{P_{{{\rm{N}}_2}{{\rm{O}}_4}}}} \right)}}{{{{\left( {{P_{{\rm{N}}{{\rm{O}}_2}}}} \right)}^2}}}\). The value of \({K_p}\) must remain the same when the pressure increases to 9.0 atm.

Determine partial pressure of each component:

b)

The reaction is\(2{\rm{N}}{{\rm{O}}_{2(g)}} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {{\rm{N}}_2}{{\rm{O}}_{4(g)}}\).

At equilibrium,

\({K_P} = \frac{{{P_{{{\rm{N}}_2}{{\rm{O}}_4}}}}}{{{{\left( {{P_{{\rm{N}}{{\rm{O}}_2}}}} \right)}^2}}} = \frac{{0.70}}{{{{(0.30)}^2}}} = \frac{{0.70}}{{0.09}} = 7.78\)

For a total pressure of 9.0 atm, the pressure of N₂O₄ is 9.0-x; that of NO is x.

\(\begin{array}{*{20}{c}}{{K_P} = \frac{{9.0 - x}}{{{x^2}}} = 7.78}\\{7.78{{\rm{x}}^2} + {\rm{x}} - 9.0 = 0}\end{array}\)

Use the quadratic expression, where

\(\begin{array}{*{20}{c}}\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{ - 1 \pm \sqrt {1 + 4(7.78)( - 9.0)} }}{{2(7.78)}}\end{array}\\\begin{array}{l} = \frac{{ - 1 \pm 16.765}}{{15.56}}\\ = 1.013\end{array}\end{array}\)

For the plus sign (the negative sign gives a negative pressure which is impossible)

\(x = \frac{{15.765}}{{15.56}} = 1.013\)

The other answer is extraneous. The pressures are \({P_{{{\rm{N}}_2}{{\rm{O}}_4}}} = 8\,atm\)and \({P_{{\rm{N}}{{\rm{O}}_2}}} = 1\,atm\).