How Do You Determine Equilibrium Concentrations in a Chemical Reaction?

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Discussion Overview

The discussion revolves around determining equilibrium concentrations in a chemical reaction, specifically for the reaction N2(g) + 3H2(g) <--> 2NH3(g) with a given equilibrium constant (K) of 64 at 25 degrees Celsius. Participants explore the setup of an equilibrium table and the application of the equilibrium constant expression.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equilibrium expression and sets up an equilibrium table, seeking to solve for 'x' to find equilibrium concentrations.
  • Another participant questions the necessity of raising equilibrium concentrations to powers in the equilibrium expression, providing an alternative formulation of the expression.
  • A third participant outlines the general form of the equilibrium expression for reactions but expresses uncertainty about the use of 'x' in the specific problem presented.
  • A later reply indicates that the original poster has resolved their question independently.

Areas of Agreement / Disagreement

Participants express differing views on the formulation of the equilibrium expression and the use of 'x' in calculations. The discussion does not reach a consensus on these points, and uncertainty remains regarding the correct approach.

Contextual Notes

There are unresolved assumptions regarding the application of stoichiometric coefficients in the equilibrium expression and the necessity of introducing 'x' in the calculations.

erjkism
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[SOLVED] Chemistry equilibrium Question

Homework Statement



K=64. at 25 degrees celsius

N2(g) + 3H2(g) <--> 2NH3(g)

Molarity N2= .5M H2= 1.5M. what are the equilibrium concentrations?

i got the equilibrium table set up, and the K expression is
(2x)(2x)
k= ------------------------------
(.5-x)(1.5-3x)(1.5-3x)(1.5-3x)

how can i solve for x?
 
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I'm not sure if you raise the equilibrium concentrations to powers; if not, I have:

K = 2x / (0.5 - x)(1.5 - 3x)
K = 2x / (3x^2 - 3x + 0.75)
(3x^2 - 3x + 0.75)K = 2x
(3x^2 - 3x + 0.75)K -2x = 0
K3x^2 - K3x + K0.75 - 2x = 0
K = 64, so:

192x^2 - 190x + 48 = 0

Plug this into the quadratic equation, then substitute the answer for x for each concentration. For example, since the equilibrium concentration for NH2 is 2x, if x = 4, the equilibrium concentration would be 8M.

Just keep in mind that I'm not sure about raising the terms to stoichiometric powers in this kind of problem...
 
For equilibrium of a reaction of the type,

aA + bB +cC + ... \leftrightharpoons\ dD + eE + ...

the equilibrium expression is:

\frac{[D]^d[E]^e...}{[A]^a<b>^b[C]^c...}</b>

I'm not sure why you are using 'x' in your problem since it really isn't necessary.
 
Last edited:
forget it i found out what to do
 

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