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### Topic: Why does the equilibrium constant expression involve coefficients  (Read 4595 times)

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#### sodium.dioxid

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##### Why does the equilibrium constant expression involve coefficients
« on: May 14, 2012, 06:28:07 PM »
Suppose we have 2NO2 N2O4
Keq=[N2O4]/[NO2]2

But the bottom term is squared. How can there be a constant ratio for two terms with different powers? (it seems that the lower term would overshoot the upper term). Why isn't it just Keq=[product]/[concentration] without the square.
« Last Edit: May 14, 2012, 06:57:44 PM by sodium.dioxid »

#### sodium.dioxid

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #1 on: May 14, 2012, 08:23:22 PM »
Suppose we have 2NO2 N2O4
Keq=[N2O4]/[NO2]2

But the bottom term is squared. How can there be a constant ratio for two terms with different powers? (it seems that the lower term would overshoot the upper term). Why isn't it just Keq=[product]/[concentration] without the square.

By the way, is this supposed to be intuitive? Or is it derived? If the latter, I will leave it alone.

#### fledarmus

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #2 on: May 14, 2012, 08:48:59 PM »
How can there be a constant ratio for two terms with different powers?

If I understand your question correctly, the answer lies in the fact that the values for the two terms are not independent. Whatever concentration of starting material you begin the reaction with absolutely fixes the final equilibrium concentration of all the components in the reaction in accordance with the equilibrium constant for the reaction.

#### sodium.dioxid

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #3 on: May 14, 2012, 09:04:25 PM »
How can there be a constant ratio for two terms with different powers?

If I understand your question correctly, the answer lies in the fact that the values for the two terms are not independent. Whatever concentration of starting material you begin the reaction with absolutely fixes the final equilibrium concentration of all the components in the reaction in accordance with the equilibrium constant for the reaction.

I understand what you are saying. You are saying that the reaction acts in such away so as to achieve concentrations that maintains the Keq. This is a mysterious conclusion. The next logical question would be why does it behave this way? How can this be physically described?

#### fledarmus

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #4 on: May 14, 2012, 09:23:54 PM »
That is where the derived part comes in. An equilibrium constant is actually a ratio of two reaction rate constants; the rate constant of the forward reaction and the rate constant of the reverse reaction. We say that a reaction is in equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction. For every two molecules of NO2 that react to form one molecule of N2O4, somewhere there is an N2O4 molecule falling apart to form two NO2 molecules.

So let's suppose that it is much faster for NO2 to react than for N2O4 to decompose - the forward rate constant is much higher than the reverse rate constant, and the Keq is a large number. For the forward reaction and the reverse reaction to occur at the same rate, you have to slow down the forward rate by making the NO2 molecules few and far between, and speed up the reverse rate by increasing the number of N2O4 molecules. At equilibrium, the concentration of product molecules is much higher than the concentration of starting material molecules.

If you try to increase the concentration of starting materials at this point by adding more NO2, you will speed up the forward rate even more and form more product, decreasing the amount of NO2 again until a new equilibrium is established, still in accord with the same Keq

#### Jorriss

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #5 on: May 14, 2012, 10:08:51 PM »
You can also derive equilibrium constants without appealing to kinetics from thermodynamics. A derivation of the equilibrium constant for ideal gases is very approachable if you take a look at Levine.

#### sodium.dioxid

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #6 on: May 14, 2012, 11:36:18 PM »
Ok, one more question. For any equilibrium reaction, starting with only reactants, are these statements true:

In the forward direction, rate of reaction decreases at a decreasing rate.
In the reverse direction, rate of reaction increases at a decreasing rate.

#### Jorriss

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #7 on: May 15, 2012, 01:17:10 AM »
Ok, one more question. For any equilibrium reaction, starting with only reactants, are these statements true:

In the forward direction, rate of reaction decreases at a decreasing rate.
In the reverse direction, rate of reaction increases at a decreasing rate.
You've taken calculus, think about this in terms of calculus. You want to know if the rate of the reaction is monotonically decreasing. How do you tell if a function is monotonically decreasing (increasing)?

#### sodium.dioxid

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #8 on: May 15, 2012, 09:25:23 AM »
Ok, one more question. For any equilibrium reaction, starting with only reactants, are these statements true:

In the forward direction, rate of reaction decreases at a decreasing rate.
In the reverse direction, rate of reaction increases at a decreasing rate.
You've taken calculus, think about this in terms of calculus. You want to know if the rate of the reaction is monotonically decreasing. How do you tell if a function is monotonically decreasing (increasing)?

Got it. For this particular problem, rate is a function of concentration squared in the forward direction. Taking the derivative of this shows a linearly increasing deceleration. So that's a check for the first one. The reverse direction is a problem. While the function is linear, the concentrations do not build up linearly. The surge comes in the beginning and fades down with increasing concentration (because the gap between the rates are closing). Nevertheless, while the gap is closing, the concentration of product is increasing, just not as fast since the difference is becoming smaller. Thus, That's a check for the second one also.

#### Jorriss

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #9 on: May 15, 2012, 11:40:32 AM »

Got it. For this particular problem, rate is a function of concentration squared in the forward direction. Taking the derivative of this shows a linearly increasing deceleration. So that's a check for the first one. The reverse direction is a problem. While the function is linear, the concentrations do not build up linearly.
Not quite, don't forget implicit diff. You differentiate with respect to time so you have to use implicit differentiation on A. y(t)=x(t)^2 => dy/dt = 2x (dx/dt)

#### sodium.dioxid

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #10 on: May 15, 2012, 06:23:25 PM »

Got it. For this particular problem, rate is a function of concentration squared in the forward direction. Taking the derivative of this shows a linearly increasing deceleration. So that's a check for the first one. The reverse direction is a problem. While the function is linear, the concentrations do not build up linearly.
Not quite, don't forget implicit diff. You differentiate with respect to time so you have to use implicit differentiation on A. y(t)=x(t)^2 => dy/dt = 2x (dx/dt)

Oops. That makes sense.
And I completely understand now the relationship between equilibrium constant and rate law constants.
But the book just gave me another surprise. In my own words, it says that the powers come from the coefficients in the composite reaction, not the powers in the rate law. Why is that?

Is it because equilibrium reactions don't occur in a composite fashion? In other words, is the composite reaction itself an elementary reaction (for equilibrium reactions)?
« Last Edit: May 15, 2012, 07:05:19 PM by sodium.dioxid »

#### sodium.dioxid

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##### Re: Why does the equilibrium constant expression involve coefficients
« Reply #11 on: May 15, 2012, 10:20:54 PM »
Please don't mind my last question. I got myself another textbook, which explained this particular topic more clearly.