Chemical Equilibria: Reversible Reactions & Dynamic Equilibrium (Cambridge (CIE) A Level Chemistry): Flashcards

The position of equilibrium shifts to the right, towards the products, to reduce the increased concentration of the reactant.

False.

Increasing pressure shifts the equilibrium towards the side with the fewer gas molecules, to reduce the pressure.

If the temperature of an equilibrium is increased, the position of equilibrium shifts in the endothermic direction to absorb the extra energy.

A catalyst has no effect on the position of equilibrium. It increases the rate of both the forward and reverse reactions equally, so equilibrium is reached faster but the equilibrium position is unchanged.

False.

Decreasing temperature shifts the equilibrium in the exothermic direction. For an exothermic forward reaction, this means shifting to the right, increasing the yield of products.

Adding water shifts the equilibrium towards the side with more particles (Le Chatelier's principle). For a weak acid equilibrium (HA ⇌ H⁺ + A^-), dilution shifts it to the right, increasing the degree of dissociation even though all ion concentrations fall.

When the concentration of a product is increased in an equilibrium system, the position of equilibrium shifts to the left to oppose the change.

The equilibrium shifts to the left, towards N₂O₄ (g), because there is one mole of gas on the left compared to two on the right, reducing the total number of gas molecules and lowering pressure.

An equilibrium constant K_c is a value that links the equilibrium concentrations of products and reactants at a given temperature, with each concentration raised to the power of its stoichiometric coefficient.

K_c = [C]^c [D]^d / [A]^a [B]^b

Concentrations of products appear on the top and reactants on the bottom, each raised to the power of their stoichiometric coefficient.

False.

Solids are excluded from the K_c expression because their concentration does not change during the reaction.

A partial pressure of a gas in a mixture is the pressure that gas would exert if it occupied the container alone. The total pressure equals the sum of all partial pressures.

The partial pressure of a gas equals its mole fraction multiplied by the total pressure.

The mole fraction of a gas is the number of moles of that gas divided by the total moles of all gases in the mixture. All mole fractions in a mixture must sum to 1.00.

True.

K_p uses partial pressures rather than concentrations, making it more practical for reactions involving gaseous mixtures.

K_p = (p NH₃)² / (p N₂) (p H₂)³

Partial pressures of products over reactants, each raised to the power of the stoichiometric coefficient.

The value of K_c only changes when the temperature of the reaction changes, not when concentration or pressure is altered.

Concentration (mol dm^-3) = number of moles / volume (dm³)

This gives the equilibrium concentration needed to substitute into the K_c expression.

False.

The units of K_c depend on the form of the equilibrium expression. If the number of moles of products and reactants are equal, all units cancel and K_c is unitless.

An ICE table (Initial, Change, Equilibrium) is used to organise mole or concentration data, using stoichiometric ratios to find unknown equilibrium values from initial and one equilibrium amount.

In a K_p calculation, the mole fraction of a gas is multiplied by the total pressure to give its partial pressure.

1. Calculate the total moles of all gases.

2. Find the mole fraction of each gas (moles / total moles).

3. Multiply each mole fraction by the total pressure to get each partial pressure.

4. Substitute partial pressures into the K_p expression.

False.

For this reaction, the total moles of gas on each side are equal (1 + 1 = 2), so all pressure units cancel and K_p is unitless.

Write Pa (or kPa) in place of each partial pressure, apply the exponents, then cancel. If more moles of gas appear on top, the units will be Pa raised to a positive power. If more appear on the bottom, the units will be Pa raised to a negative power.

When constructing an ICE table, a decrease in moles is shown as negative and an increase is shown as positive in the change row.

K_p = Pa² / (Pa² x Pa) = Pa^-1

There are two moles of gas on top and three on the bottom, so the units simplify to Pa^-1.

Only a change in temperature changes the value of K_c. Changes in concentration, pressure and the addition of a catalyst do not affect K_c.

False.

Adding more reactant shifts the equilibrium position to the right, but the ratio of products to reactants adjusts back to the same value. K_c remains unchanged.

For an endothermic reaction, increasing the temperature causes K_c to increase because the concentration of products increases relative to reactants.

A catalyst speeds up both the forward and reverse reactions by the same factor, so the ratio of product concentrations to reactant concentrations at equilibrium is unchanged and K_c stays the same.

False.

For an exothermic reaction, increasing temperature shifts equilibrium to the left, decreasing product concentrations and increasing reactant concentrations. K_c decreases.

An equilibrium position is the relative distribution of reactants and products in an equilibrium mixture. It can shift left or right in response to changes in concentration, pressure or temperature.

K_c decreases. The forward reaction is exothermic, so increasing temperature shifts equilibrium to the left. [SO₃] decreases and [SO₂] and [O₂] increase, making the K_c ratio smaller.

A change in pressure shifts the position of equilibrium but does not change the value of K_c, as long as temperature remains constant.

Increasing temperature would increase K_p. The forward reaction is endothermic, so higher temperature shifts equilibrium to the right, increasing product partial pressures and raising K_p.

The Haber process is the industrial synthesis of ammonia from nitrogen and hydrogen: N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g), ΔH_r = −92 kJ mol⁻¹.

The forward reaction is exothermic, so lower temperatures would give a higher equilibrium yield of NH₃. However, too low a temperature makes the reaction rate very slow, so 400–450 °C is a compromise between yield and rate.

False.

The iron catalyst increases the rate of reaction, allowing equilibrium to be reached faster. It has no effect on the equilibrium position or the yield of ammonia.

In the Haber process, ammonia is removed by cooling and condensing it, which shifts the equilibrium to the right and increases the yield.

At low pressures the equilibrium constant K_p for the Contact process is already very large, so the yield of SO₃ is high. Increasing pressure would add cost with little extra benefit.

True.

Vanadium(V) oxide (V₂O₅) is the catalyst used in the Contact process to increase the rate of SO₂ oxidation without affecting the equilibrium position.

SO₃ is absorbed into 98% H₂SO₄ to form oleum (H₂S₂O₇). Removing SO₃ shifts the equilibrium to the right, driving further production of SO₃.

In the Haber process, a pressure of approximately 200 atm is used as a compromise between yield and the cost of maintaining high pressure.

The reaction is exothermic, so lower temperatures favour SO₃ production, but they also reduce the reaction rate. 450 °C is a compromise that gives an acceptable rate while still achieving a good yield.