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Pressure and Chemical Equilibrium

discuss the e...

Question

discuss the effect of catalyst and addingof inert gas at constant pressure for the reaction.
2so2(g)+o2(g)=2so3(g)

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Solution

2SO₂(g) + O₂(g) $\leftrightarrow$ 2SO₃(g)

a) Effect of catalyst: A catalyst increases the rate of forward and the backward reaction to the same extent. This is due to the fact that it decreases the energy of activation of the forward as well as backward reaction to the same extent. Hence a catalyst has no effect on the final state of equilibrium. A catalyst, however, enables the system to reach a state of equilibrium much quickly.

b) Addition of inert gas at constant pressure : The addition of inert gas at constant pressure will increase the total volume. Hence, at equilibrium the molar concentration of reactants and products decrease and the equilibrium will shift in a direction in which number of moles increase. Therefore, in above reaction, equilibrium shifts towards left.

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Q. The equilibrium constant

K_{p}

for the reaction

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

1000 K

is 3.5. What would be the partial pressure of

O_{2}

gas with equal moles of

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S O_{3}

at equilibrium?

Q. The equilibrium constant for the reaction:

2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}

is 5. If the equilibrium constant mixture contains equal moles of

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Q. For the reaction,

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

300 K

Δ G^{o}

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The equilibrium constant value for the reaction at that temperature is (R is gas constant)

Q. Consider the following equilibrium system;

2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}

; some inert gas is added to the system at constant volume. Predict which of the following is true?

Q. The equilibrium constant

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2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

900 {a t m}^{- 1}

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respectively is heated at constant volume to equilibriate. Calculate the partial pressure of each gas at

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