# Thermodynamics equilibrium constant problem

• Samuel1321
In summary, the problem involves calculating the equilibrium constant at 500K and estimating the percentage conversion at equilibrium at 1 bar total pressure assuming a 1:3 stoichiometric ratio of N2 to H2. The Van Hoff equation and Dalton's law of partial pressures are used to solve for the initial and final partial pressures. An ICE table is then used to calculate the final partial pressures and mole fractions at equilibrium.
Samuel1321

## Homework Statement

N2 + 3H2 --> 2NH3 (all gases)

Why is such a high pressure needed? Calculate the equilibrium constant at 500k then estimate the percentage conversion at equilibrium at 1 bar total pressure, assuming the stoichiometric ratio of N22 is 1:3.

Repeat the process at 50 bar.

## Homework Equations

Van Hoff equation : ln(keq) at final temp = -((delta G)/RTi) - ((delta H)/R)(1/Tf-1/Ti)

Keq = ((PNH3)^2)/((PH2)^3)(PN2)

PNH3 + PH2 + PN2 = 1 bar

## The Attempt at a Solution

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I solved for keq at 500k. I have no idea how to estimate the percentage conversion at equilibrium at 1 bar total pressure since we aren't given any of the partial pressures.

I tried using molar ratios to solve but it doesn't really make sense. Here's what I did 1x + 3x + 2x = 1, 6x=1 x=0.33 = 33%. For 50 bar 1x+3x+2x=50, then 6x=50, x would be over 100%.

What are the initial partial pressures at 1 bar? What are the initial partial pressures at 50 bars?

Chestermiller said:
What are the initial partial pressures at 1 bar? What are the initial partial pressures at 50 bars?

They did not say what the initial pressures are.

Samuel1321 said:
They did not say what the initial pressures are.
You know the total pressure and the mole fractions of H2 and N2. So, from Dalton's law of partial pressures, what are the initial partial pressures of H2 and N2?

Chestermiller said:
You know the total pressure and the mole fractions of H2 and N2. So, from Dalton's law of partial pressures, what are the initial partial pressures of H2 and N2?

So assuming that there is no product in at the beginning of the reaction, partial pressure of N2 would be 1/3 bar and the partial pressure of H2 would be 2/3 bar.

Samuel1321 said:
So assuming that there is no product in at the beginning of the reaction, partial pressure of N2 would be 1/3 bar and the partial pressure of H2 would be 2/3 bar.
No, 1/4 and 3/4. These are the mole fractions.

Samuel1321
Chestermiller said:
No, 1/4 and 3/4. These are the mole fractions.

Oh right, just had a brain fart, don't know how I missed that. So how should I proceed? Should I set up an equation for Keq?

Let x be the fraction of the reactants converted to ammonia. What are the new mole fractions of N2 and H2, and what is the mole fraction ammonia.

Chestermiller said:
Let x be the fraction of the reactants converted to ammonia. What are the new mole fractions of N2 and H2, and what is the mole fraction ammonia.
Thanks for the reply! I appreciate the help! I asked my professor and I figured it out.

## 1. What is thermodynamic equilibrium constant?

Thermodynamic equilibrium constant is a measure of the ratio of products to reactants at equilibrium for a specific chemical reaction. It is denoted by the symbol K and is dependent on temperature.

## 2. How is thermodynamic equilibrium constant calculated?

Thermodynamic equilibrium constant is calculated by taking the ratio of the concentration of products to the concentration of reactants, each raised to their respective stoichiometric coefficient, at equilibrium. The concentrations are typically in molarity (M) and the value of K is unitless.

## 3. What factors affect the value of thermodynamic equilibrium constant?

The value of thermodynamic equilibrium constant is affected by temperature, pressure, and the initial concentrations of reactants and products. Changes in any of these factors can alter the value of K for a chemical reaction.

## 4. How does the value of thermodynamic equilibrium constant relate to the direction of a chemical reaction?

The value of thermodynamic equilibrium constant determines the direction of a chemical reaction at equilibrium. If K is greater than 1, the reaction favors the formation of products and if K is less than 1, the reaction favors the formation of reactants. A value of 1 indicates that the reaction is at equilibrium.

## 5. Can the value of thermodynamic equilibrium constant change?

Yes, the value of thermodynamic equilibrium constant can change if the temperature, pressure, or concentrations of reactants and products change. This can result in a shift in the equilibrium position and a change in the direction of the reaction.

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