Consider the following system at equilibrium: NO(g)+SO3(g) ⟹NO2(g)+SO2(g) Indicate the effect of

Consider the following system at equilibrium: NO(g)+SO3(g)...

Consider the following system at equilibrium: $$ \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \Longrightarrow \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) $$ Indicate the effect of the following changes. (a) an increase in the concentration of NO (b) an increase in the concentration of $\mathrm{SO}_{3}$ (c) a decrease in the concentration of $\mathrm{NO}_{2}$ (d) an increase in the concentration of $\mathrm{NO}_{2}$

Consider the following system at equilibrium:
$$
\mathrm{NO}(g)+\mathrm{SO}_{3}(g) \Longrightarrow \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g)
$$
Indicate the effect of the following changes.
(a) an increase in the concentration of NO
(b) an increase in the concentration of $\mathrm{SO}_{3}$
(c) a decrease in the concentration of $\mathrm{NO}_{2}$
(d) an increase in the concentration of $\mathrm{NO}_{2}$

Key Concepts

Chemical Equilibrium

Chemical equilibrium is the state of a reversible reaction where the forward and reverse reaction rates are equal, resulting in constant concentrations of reactants and products over time. This balance is dynamic, meaning that even though no net change is observed, both reactions continue to occur.

Le Chatelier's Principle

Le Chatelier's Principle explains how a system at equilibrium responds to disturbances such as changes in concentration, pressure, or temperature. When a change is imposed on the system, it shifts in a direction that opposes that change by either producing more reactants or products, thereby re-establishing equilibrium.

Impact of Concentration Changes on Equilibrium

Changes in the concentration of reactants or products disturb the equilibrium balance. An increase in the concentration of reactants typically shifts the equilibrium toward the products, while an increase in the concentration of products shifts it toward the reactants. Conversely, decreasing the concentration of either side causes the system to shift to replace what was removed.

Transcript

00:01 We are given this reaction and we want to determine how the equilibrium position will change when we impose the various disturbances on the system outlined in parts a through c.

00:13 In part a, we want to determine how the system will respond if we increase the total pressure by decreasing the volume.

00:22 According to the shatlier's principle, if the stress that we impose on the system is an increase in pressure, then the system will receive.

00:31 Spawn to get back to equilibrium by decreasing the pressure.

00:35 The way that we decrease the pressure for a gas phase reaction is to form the fewest number of moles of gas as possible.

00:43 And the way that we do that is by favoring the side that has the fewer number of moles.

00:49 We see that we have two moles of gas on the reactant side and one mole of gas on the product side.

00:55 And so in order to decrease the number of moles of gas, we have to favor the right side, the product i2, in order to decrease the overall pressure.

01:06 Because when we have fewer number of moles of gas, then we have fewer contributions to the total pressure based on the partial pressures of those gases.

01:17 And so that is why in part a, the equilibrium position will shift to the right.

01:26 And in part b, we want to determine the effect of adding i2 so increasing the concentration of i2 gas for the effect of concentration we should consider the equilibrium expression for kc that would be equilibrium concentration of i2 divided by the equilibrium concentration of i squared and this is also this expression is also equal to q, but these are general concentrations where when we take the ratio, it no longer comes out to the value of the equilibrium constant because both of these concentrations no longer correspond to their equilibrium concentrations because we changed the concentration of one or both of these species to get it out of equilibrium.

02:21 And in part b, we specifically increased the concentration of i2 gas from its equilibrium value.

02:32 And so since that appears on the numerator, when we take their ratio for q, we would find that q is greater than k.

02:40 And whenever q is greater than k, the reaction will respond by shifting to the left to form more reactants.

02:48 So when we add i2, we increase its concentration, and we have to form more reactants...

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