Animated Solution for Chemistry - Chemical Equilibrium: The exothermic formation of ${\text{ClF}}_3$ is represented by the equation ${\text{Cl}}_2(g)+3{\text{F}}_2(g)\rightleftharpoons 2{\text{ClF}}_3(g)$; $\Delta H_r=-329\text{ kJ}$ Which of the following will increase the quantity of ${\text{ClF}}_3$ in an equilibrium mixture of ${\text{Cl}}_2$, ${\text{F}}_2$ and ${\text{ClF}}_3$?

Imagine a perfectly balanced seesaw. On one side, you have your reactants, and on the other, your products. When a chemical reaction reaches equilibrium, it's like that perfectly still seesaw. The forward and backward reactions are happening at the exact same rate, so the overall amounts of reactants and products don't change.

But what happens if you suddenly drop a heavy weight on one side? The seesaw tips. In chemistry, this 'tipping' is governed by a beautiful concept known as Le Chatelier's Principle. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. The system acts like a stubborn toddler—whatever you do to it, it tries to undo.

Let's apply this to our specific reaction:

We are told that the enthalpy change, $\Delta H_r$, is $-329\text{ kJ}$. The negative sign is a massive clue: it means the reaction is exothermic. It releases heat as it moves forward. Our goal is to figure out how to trick this system into making more of our product, ${\text{ClF}}_3$.

Let's meticulously break down each option to see how the stubborn system will react.

Option (a): Adding ${\text{F}}_2$ If we pump more fluorine gas (${\text{F}}_2$) into the container, we are suddenly increasing the concentration of a reactant. The system looks around and says, "Whoa, there's way too much fluorine here!" To counteract this addition, it must consume the excess fluorine. How does it do that? By reacting it with chlorine to form more product. Therefore, the equilibrium shifts in the forward direction, increasing the quantity of ${\text{ClF}}_3$. This looks like our winner!

Option (b): Increasing the volume of the container When you increase the volume of a gas container, the pressure drops. The system hates this drop in pressure and wants to bring it back up. It does this by shifting towards the side of the reaction that has more moles of gas.

Let's count the gaseous moles: Reactants: $1\text{ mole of }{\text{Cl}}_2+3\text{ moles of }{\text{F}}_2=4\text{ moles}$ Products: $2\text{ moles of }{\text{ClF}}_3=2\text{ moles}$

Since $4>2$, the system will shift backward to create more gas molecules and raise the pressure. This decreases our product yield.

Option (c): Removing ${\text{Cl}}_2$ If we extract chlorine gas from the mixture, the system feels a shortage. To replace the lost ${\text{Cl}}_2$, it will break down the ${\text{ClF}}_3$ we worked so hard to make. The equilibrium shifts backward, which is the exact opposite of what we want.

Option (d): Increasing the temperature Remember that $\Delta H_r=-329\text{ kJ}$? Because the forward reaction is exothermic, you can think of 'Heat' as a product:

If we increase the temperature, we are essentially adding more 'Heat' to the product side. The system will try to consume this excess heat by shifting in the endothermic direction, which is backward. Again, our product yield drops.

By systematically applying Le Chatelier's Principle, we can confidently conclude that the only way to force the system to produce more ${\text{ClF}}_3$ is by feeding it more reactants. Adding ${\text{F}}_2$ creates a forward shift, making Option (a) the correct answer.