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Some questions have been on my mind for a long time, but I’ve never been able to find satisfactory answers. I’ve read some possible duplicates to my question, but only a few of my concerns were addressed. While answers to similar questions exist, they often include advanced formulas and terms that I, as a beginner studying chemical equilibrium, find difficult to understand. Hence, I’m unable to grasp the major parts of those explanations.

$$\mathrm{[activity~coefficient (\lambda)×[conc]~or~ [Partial Pressure]}$$

and mentioned that, for the scope of our course, $\lambda=1.$

More specifically, what qualifies as "impure" in this context? I’ve gone through several textbooks, but none explicitly discuss this part.

Some books address this question with thermochemical or mathematical proofs, but I find them hard to follow at my current level of understanding. I did come across an explanation for why solids are excluded from the equilibrium constant that gave me a good intuitive understanding:

More solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsely granulated sugar, but the solubility (and the equilibrium constant) remains the same.

This explanation helped me understand solids better, but it does not provide a similar intuition for liquids. Could someone provide a comparable intuition for liquids?

Is the equilibrium constant $(K_{C})$ correctly written as: $$K_{C}=\frac{ \ce{[OH-] [H3O+]}}{ \ce{[H2O]^2}}=\frac{ \ce{[OH-][H3O+]}}{1}= \ce{[OH-][H3O^+]} ?$$

I've often seen the statement that in dilute solutions, $\ce{[H2O]}$ is treated as constant, but I’m unsure why this is the case. Since water remains a liquid in both dilute and concentrated solutions, shouldn’t its active mass be considered constant in both cases? Why is the assumption about water's concentration only made for dilute solutions?

Note: I know StackExchange restricts asking multiple questions within one post, but it also restricts asking more than two questions within a given time. If I posted the questions one by one, then the synchronization I want between the questions will not be there. But I am still sorry to do it.

$$\mathrm{[activity~coefficient (\lambda)×[conc]~or~ [Partial~Pressure]}$$

and mentioned that, for the scope of our course, $\lambda=1$.

More specifically, what qualifies as "impure" in this context? I’ve gone through several textbooks, but none explicitly discuss this part.

Some books address this question with thermochemical or mathematical proofs, but I find them hard to follow at my current level of understanding. I did come across an explanation for why solids are excluded from the equilibrium constant that gave me a good intuitive understanding:

More solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsely granulated sugar, but the solubility (and the equilibrium constant) remains the same.

This explanation helped me understand solids better, but it does not provide a similar intuition for liquids. Could someone provide a comparable intuition for liquids?

Is the equilibrium constant $(K_{C})$ correctly written as: $$K_{C}=\frac{ \ce{[OH-] [H3O+]}}{ \ce{[H2O]^2}}=\frac{ \ce{[OH-][H3O+]}}{1}= \ce{[OH-][H3O+]} ?$$

I've often seen the statement that in dilute solutions, $\ce{[H2O]}$ is treated as constant, but I’m unsure why this is the case. Since water remains a liquid in both dilute and concentrated solutions, shouldn’t its active mass be considered constant in both cases? Why is the assumption about water's concentration only made for dilute solutions?

Some questions have been on my mind for a long time, but I’ve never been able to find satisfactory answers. I’ve read some possible duplicates to my question, but only a few of my concerns were addressed. While answers to similar questions exist, they often include advanced formulas and terms that I, as a beginner studying chemical equilibrium, find difficult to understand. Hence, I’m unable to grasp the major parts of those explanations.

Background

The professor who taught us chemical equilibrium defined active mass as:

$$\mathrm{[activity~coefficient (\lambda)×[conc]~or~ [Partial Pressure]}$$

and mentioned that, for the scope of our course, $\lambda=1.$

Questions

1. What do "pure solids" and "pure liquids" mean in this context?

More specifically, what qualifies as "impure" in this context? I’ve gone through several textbooks, but none explicitly discuss this part.

2. Why do pure solids and pure liquids have an active mass equal to 1?

Some books address this question with thermochemical or mathematical proofs, but I find them hard to follow at my current level of understanding. I did come across an explanation for why solids are excluded from the equilibrium constant that gave me a good intuitive understanding:

More solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsely granulated sugar, but the solubility (and the equilibrium constant) remains the same.

This explanation helped me understand solids better, but it does not provide a similar intuition for liquids. Could someone provide a comparable intuition for liquids?

3. Equilibrium constant expressions

For the reaction:

$$\ce{2 H2O⇋H3O+ + OH-}$$

Is the equilibrium constant $(K_{C})$ correctly written as: $$K_{C}=\frac{[OH^{-}]^{1}[H_{3}O^{+}]^{1}}{[H_{2}O]^{2}}=\frac{[OH^{-}][H_{3}O^{+}]}{1}=[OH^{-}][H_{3}O^{+}] ?$$

And, for the hypothetical reaction $$\ce{2A(l)+4C(s)⇋1B(g)+2D(s)+3E(aq)}$$ $$K_{C}=\frac{[B]^{1}[D]^{2}[E]^{3}}{[A]^{2}[C]^{4}}=[B][E]^{3} ?$$

4. Why is it said that $\ce{[H2O]}$ is approximately constant in dilute solutions but not in concentrated solutions?

I've often seen the statement that in dilute solutions, $\ce{[H2O]}$ is treated as constant, but I’m unsure why this is the case. Since water remains a liquid in both dilute and concentrated solutions, shouldn’t its active mass be considered constant in both cases? Why is the assumption about water's concentration only made for dilute solutions?

Note: I know StackExchange restricts asking multiple questions within one post, but it also restricts asking more than two questions within a given time. If I posted the questions one by one, then the synchronization I want between the questions will not be there. But I am still sorry to do it.

Some questions have been on my mind for a long time, but I’ve never been able to find satisfactory answers. I’ve read some possible duplicates to my question, but only a few of my concerns were addressed. While answers to similar questions exist, they often include advanced formulas and terms that I, as a beginner studying chemical equilibrium, find difficult to understand. Hence, I’m unable to grasp the major parts of those explanations.

Background

The professor who taught us chemical equilibrium defined active mass as:

$$\mathrm{[activity~coefficient (\lambda)×[conc]~or~ [Partial Pressure]}$$

and mentioned that, for the scope of our course, $\lambda=1.$

Questions

1. What do "pure solids" and "pure liquids" mean in this context?

More specifically, what qualifies as "impure" in this context? I’ve gone through several textbooks, but none explicitly discuss this part.

2. Why do pure solids and pure liquids have an active mass equal to 1?

Some books address this question with thermochemical or mathematical proofs, but I find them hard to follow at my current level of understanding. I did come across an explanation for why solids are excluded from the equilibrium constant that gave me a good intuitive understanding:

More solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsely granulated sugar, but the solubility (and the equilibrium constant) remains the same.

This explanation helped me understand solids better, but it does not provide a similar intuition for liquids. Could someone provide a comparable intuition for liquids?

3. Equilibrium constant expressions

For the reaction:

$$\ce{2 H2O⇋H3O+ + OH-}$$

Is the equilibrium constant $(K_{C})$ correctly written as: $$K_{C}=\frac{ \ce{[OH-] [H3O+]}}{ \ce{[H2O]^2}}=\frac{ \ce{[OH-][H3O+]}}{1}= \ce{[OH-][H3O^+]} ?$$

And, for the hypothetical reaction $$\ce{2A_(l) + 4C_(s) ⇋ B_(g) + 2D_(s) + 3E_(aq)}$$ $$K_{C}=\frac{ \ce{[B]^{1}[D]^{2}[E]^{3}}}{ \ce{[A]^{2}[C]^{4}}}= \ce{[B][E]^{3}} ?$$

4. Why is it said that $\ce{[H2O]}$ is approximately constant in dilute solutions but not in concentrated solutions?

I've often seen the statement that in dilute solutions, $\ce{[H2O]}$ is treated as constant, but I’m unsure why this is the case. Since water remains a liquid in both dilute and concentrated solutions, shouldn’t its active mass be considered constant in both cases? Why is the assumption about water's concentration only made for dilute solutions?

Note: I know StackExchange restricts asking multiple questions within one post, but it also restricts asking more than two questions within a given time. If I posted the questions one by one, then the synchronization I want between the questions will not be there. But I am still sorry to do it.

Some questions have been on my mind for a long time, but I’ve never been able to find satisfactory answers. I’ve read some possible duplicates to my question, but only a few of my concerns were addressed. While answers to similar questions exist, they often include advanced formulas and terms that I, as a beginner studying chemical equilibrium, find difficult to understand. Hence, I’m unable to grasp the major parts of those explanations.


 

Background

The professor who taught us Chemical Equilibrium defined active mass as:$$activity~coefficient (\lambda)×[conc]~or~ [Partial Pressure]$$

and mentioned that, for the scope of our course, $\lambda=1.$

Questions

1. What do "pure solids" and "pure liquids" mean in this context?

More specifically, what qualifies as "impure" in this context? I’ve gone through several textbooks, but none explicitly discuss this part.

2. Why do pure solids and pure liquids have an active mass equal to 1?

Some books address this question with thermochemical or mathematical proofs, but I find them hard to follow at my current level of understanding. I did come across an explanation for why solids are excluded from the equilibrium constant that gave me a good intuitive understanding:

More solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsely granulated sugar, but the solubility (and the equilibrium constant) remains the same.

This explanation helped me understand solids better, but it does not provide a similar intuition for liquids. Could someone provide a comparable intuition for liquids?

3. Equilibrium constant expressions

For the reaction: $$2H_{2}O⇋H_{3}O^{+}+OH^{-}$$ Is the equilibrium constant $(K_{C})$ correctly written as: $$K_{C}=\frac{[OH^{-}]^{1}[H_{3}O^{+}]^{1}}{[H_{2}O]^{2}}=\frac{[OH^{-}][H_{3}O^{+}]}{1}=[OH^{-}][H_{3}O^{+}] ?$$

And, for the hypothetical reaction $$2A(l)+4C(s)⇋1B(g)+2D(s)+3E(aq)$$ $$K_{C}=\frac{[B]^{1}[D]^{2}[E]^{3}}{[A]^{2}[C]^{4}}=[B][E]^{3} ?$$

4. Why is it said that $[H_{2}O]$ is approximately constant in dilute solutions but not in concentrated solutions?

I've often seen the statement that in dilute solutions, $[H_{2}O]$ is treated as constant, but I’m unsure why this is the case. Since water remains a liquid in both dilute and concentrated solutions, shouldn’t its active mass be considered constant in both cases? Why is the assumption about water's concentration only made for dilute solutions?

Note: I know StackExchange restricts asking multiple questions within one post, but it also restricts asking more than two questions within a given time. If I posted the questions one by one, then the synchronization I want between the questions will not be there. But I am still sorry to do it.

Thank you for taking the time to read through my lengthy question. I truly appreciate your patience and any insights you can provide to help clarify these points!

Some questions have been on my mind for a long time, but I’ve never been able to find satisfactory answers. I’ve read some possible duplicates to my question, but only a few of my concerns were addressed. While answers to similar questions exist, they often include advanced formulas and terms that I, as a beginner studying chemical equilibrium, find difficult to understand. Hence, I’m unable to grasp the major parts of those explanations.

Background

The professor who taught us chemical equilibrium defined active mass as:

$$\mathrm{[activity~coefficient (\lambda)×[conc]~or~ [Partial Pressure]}$$

and mentioned that, for the scope of our course, $\lambda=1.$

Questions

1. What do "pure solids" and "pure liquids" mean in this context?

More specifically, what qualifies as "impure" in this context? I’ve gone through several textbooks, but none explicitly discuss this part.

2. Why do pure solids and pure liquids have an active mass equal to 1?

Some books address this question with thermochemical or mathematical proofs, but I find them hard to follow at my current level of understanding. I did come across an explanation for why solids are excluded from the equilibrium constant that gave me a good intuitive understanding:

More solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsely granulated sugar, but the solubility (and the equilibrium constant) remains the same.

This explanation helped me understand solids better, but it does not provide a similar intuition for liquids. Could someone provide a comparable intuition for liquids?

3. Equilibrium constant expressions

For the reaction:

$$\ce{2 H2O⇋H3O+ + OH-}$$

Is the equilibrium constant $(K_{C})$ correctly written as: $$K_{C}=\frac{[OH^{-}]^{1}[H_{3}O^{+}]^{1}}{[H_{2}O]^{2}}=\frac{[OH^{-}][H_{3}O^{+}]}{1}=[OH^{-}][H_{3}O^{+}] ?$$

And, for the hypothetical reaction $$\ce{2A(l)+4C(s)⇋1B(g)+2D(s)+3E(aq)}$$ $$K_{C}=\frac{[B]^{1}[D]^{2}[E]^{3}}{[A]^{2}[C]^{4}}=[B][E]^{3} ?$$

4. Why is it said that $\ce{[H2O]}$ is approximately constant in dilute solutions but not in concentrated solutions?

I've often seen the statement that in dilute solutions, $\ce{[H2O]}$ is treated as constant, but I’m unsure why this is the case. Since water remains a liquid in both dilute and concentrated solutions, shouldn’t its active mass be considered constant in both cases? Why is the assumption about water's concentration only made for dilute solutions?

Note: I know StackExchange restricts asking multiple questions within one post, but it also restricts asking more than two questions within a given time. If I posted the questions one by one, then the synchronization I want between the questions will not be there. But I am still sorry to do it.